Chapter 8 · Application of Integrals

8.2 Areas Bounded by Lines, Parabolas, Circles and Ellipses

Straight lines give triangles and trapeziums; the parabola $y^2=4ax$ gives regions that are exactly two-thirds of their bounding rectangles; the circle gives $\pi a^2$; and stretching the circle gives the ellipse area $\pi ab$ — all from one integral and a little symmetry.

An Earth-observation satellite launched by ISRO does not circle the Earth: it moves on an ellipse, with the Earth at one focus. Four hundred years ago Kepler discovered that the line from the Earth (for him, the Sun) to the satellite sweeps out equal areas in equal times. So if you know the area of the whole ellipse and the time for one orbit, you know how much area is swept every minute — and with that, where the satellite will be. All of this needs one formula: the area of an ellipse with semi-axes $a$ and $b$. Is it $\pi ab$? Why should it be?

In Section 8.1 you learnt the general method — slice the region into strips and integrate. This section applies it to the curves you know from Class 11 Mathematics, Ch. 10 — Conic Sections, in their standard positions: straight lines, the parabolas $y^2=4ax$ and $x^2=4ay$, the circle $x^2+y^2=a^2$ and the ellipse $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$, together with an axis or a line parallel to an axis. Each region in this section is bounded by one such curve and straight lines parallel to the axes (or an axis itself). Along the way you will rediscover Archimedes' two-thirds rule for parabolas and prove the formulas $\pi a^2$ and $\pi ab$. The key tool, besides the integral, is symmetry: each of these curves is symmetric about an axis, so you can find one quarter or one half of the region and multiply.

Regions Bounded by Straight Lines

For a straight line you could find every area with the triangle and trapezium formulas — which makes lines the ideal place to test the integral method. Two points need care: a line usually crosses the $x$-axis somewhere, so split there (Theorem 8.3); and a line through the axes can be integrated with either vertical or horizontal strips.

✏️ Example 1 — Lines and the axes

(a) Find the area of the region bounded by the line $y=3x+2$, the $x$-axis and the lines $x=-1$ and $x=1$. (b) Find the area of the triangle bounded by the line $2x+3y=12$ and the coordinate axes, using horizontal strips.

💡 Show solution▶

(a) sketch $3x+2=0$ at $x=-\tfrac23$. The line is below the axis on $\left(-1,-\tfrac23\right)$ and above it on $\left(-\tfrac23,1\right)$, so split at $-\tfrac23$. With $F(x)=\tfrac32x^2+2x$: $F(-1)=-\tfrac12$, $F\!\left(-\tfrac23\right)=-\tfrac23$, $F(1)=\tfrac72$.

(a) pieces $\displaystyle\int_{-1}^{-2/3}(3x+2)\,dx=-\tfrac23+\tfrac12=-\tfrac16$ and $\displaystyle\int_{-2/3}^{1}(3x+2)\,dx=\tfrac72+\tfrac23=\tfrac{25}6$. Area $=\tfrac16+\tfrac{25}6=\tfrac{13}3$ square units.

(a) check Two triangles: base $\tfrac13$, height $1$ (area $\tfrac16$) and base $\tfrac53$, height $5$ (area $\tfrac{25}6$) ✓. Without splitting you would get $\tfrac{25}6-\tfrac16=4$ — wrong.

(b) The line meets the axes at $(6,0)$ and $(0,4)$. A horizontal strip at height $y$ runs from the $y$-axis to the line, where $x=\dfrac{12-3y}{2}$: $$A=\int_0^4\dfrac{12-3y}{2}\,dy=\dfrac12\Big[12y-\tfrac32y^2\Big]_0^4=\dfrac12(48-24)=12\ \text{square units}.$$ Check: $\tfrac12\times6\times4=12$ ✓.

Parabolas: the Two-Thirds Rule

The parabola $y^2=4ax$ ($a\gt0$) has its vertex at the origin, opens to the right and is symmetric about the $x$-axis: if $(x,y)$ is on it, so is $(x,-y)$. Its upper half is $y=2\sqrt{ax}$ and its lower half $y=-2\sqrt{ax}$. A vertical line $x=c$ with $c\gt0$ cuts off a region between the vertex and the line. In the same way the parabola $x^2=4ay$ opens upwards, is symmetric about the $y$-axis, and a horizontal line $y=c$ cuts off a region above the vertex.

📘 Definition 8.2 — Segments of a parabola and of a circle

A parabolic segment is the region bounded by a parabola and a straight line that cuts it in two points. In this chapter the line is perpendicular to the axis of the parabola — for example the region bounded by $y^2=4ax$ and the line $x=c$ ($c\gt0$). When $c=a$ the line is the latus rectum, the chord through the focus $(a,0)$ perpendicular to the axis; it has length $4a$.

A segment of a circle is the region bounded by the circle and a chord. The smaller of the two pieces is the minor segment and the larger the major segment.

📐 Theorem 8.4 — Area of a parabolic segment (the two-thirds rule)

For $a\gt0$ and $c\gt0$, the area of the region bounded by the parabola $y^2=4ax$ and the line $x=c$ is $$A=\dfrac83\sqrt a\;c^{3/2}=\dfrac23\times\big(\text{area of the rectangle } 0\le x\le c,\ -2\sqrt{ac}\le y\le2\sqrt{ac}\big).$$ In particular, the region bounded by the parabola and its latus rectum $x=a$ has area $\dfrac83a^2$. The same holds for $x^2=4ay$ and the line $y=c$, with $x$ and $y$ interchanged.

Proof. By symmetry about the $x$-axis, $A$ is twice the area of the part above the axis, which lies under $y=2\sqrt{ax}\ge0$ for $0\le x\le c$. By Theorem 8.1, $$A=2\int_0^c2\sqrt a\,x^{1/2}\,dx=4\sqrt a\cdot\dfrac23\Big[x^{3/2}\Big]_0^c=\dfrac83\sqrt a\,c^{3/2}.$$ The line $x=c$ meets the parabola at $y=\pm2\sqrt{ac}$, so the rectangle has width $c$, height $4\sqrt{ac}$ and area $4\sqrt a\,c^{3/2}$, and $\tfrac83\div4=\tfrac23$. For $c=a$, $A=\tfrac83\sqrt a\,a^{3/2}=\tfrac83a^2$. $\blacksquare$

✏️ Example 2 — The parabola and its latus rectum

Find the area of the region bounded by the parabola $y^2=12x$ and its latus rectum, working from the integral (not just quoting Theorem 8.4).

💡 Show solution▶

Identify $y^2=12x=4\cdot3\cdot x$, so $a=3$: focus $(3,0)$, latus rectum $x=3$, meeting the parabola at $(3,\pm6)$.

Symmetry The region is symmetric about the $x$-axis; the upper half lies under $y=\sqrt{12x}=2\sqrt3\,\sqrt x$ for $0\le x\le3$.

Integrate $$A=2\int_0^3 2\sqrt3\,x^{1/2}\,dx=4\sqrt3\cdot\dfrac23\Big[x^{3/2}\Big]_0^3=\dfrac{8\sqrt3}{3}\cdot3\sqrt3=24\ \text{square units}.$$

Check $\tfrac83a^2=\tfrac83\cdot9=24$ ✓, and the rectangle $0\le x\le3$, $-6\le y\le6$ has area $36$, of which $24$ is two-thirds ✓.

✏️ Example 3 — Vertical and horizontal strips on parabolas

(a) Find the area of the region in the first quadrant bounded by $y^2=9x$, the $x$-axis and the lines $x=2$ and $x=4$. (b) Find the area of the region bounded by the parabola $x^2=4y$ and the line $y=4$.

💡 Show solution▶

(a) In the first quadrant $y=3\sqrt x$. $$A=\int_2^4 3\sqrt x\,dx=2\Big[x^{3/2}\Big]_2^4=2\big(8-2\sqrt2\big)=16-4\sqrt2\approx10.34\ \text{square units}.$$

(b) set up $x^2=4y$ opens upwards; $y=4$ meets it at $x=\pm4$. The region lies between the two branches $x=\pm2\sqrt y$ for $0\le y\le4$ and is symmetric about the $y$-axis. Use horizontal strips for the right half: length $x=2\sqrt y$.

(b) integrate $$A=2\int_0^4 2\sqrt y\,dy=4\cdot\dfrac23\Big[y^{3/2}\Big]_0^4=\dfrac83\cdot8=\dfrac{64}3\ \text{square units}.$$

(b) check Theorem 8.4 with $a=1$, $c=4$: $\tfrac83\cdot1\cdot8=\tfrac{64}3$ ✓; and the rectangle $-4\le x\le4$, $0\le y\le4$ has area $32$, two-thirds of which is $\tfrac{64}3$ ✓.

Choose a region and move the cutting line with the slider. Play sweeps a red strip across the region, filling it in blue; each strip runs across the whole region, from the lower branch to the upper one (or left to right), which is why every integral below carries a factor $2$ from symmetry. The dashed box is the bounding rectangle — for the parabolas, notice the fraction that never changes.

Region:$c$:

integral—

swept so far—

exact area—

area ÷ bounding box—

Circles: Why the Area Is $\pi a^2$

You have used "area of a circle $=\pi r^2$" since middle school, usually without proof. With integration we can prove it. The circle $x^2+y^2=a^2$ has its centre at the origin and is symmetric about both axes, so its area is four times the area of the quarter in the first quadrant. That quarter lies under the upper semicircle $y=\sqrt{a^2-x^2}$ for $0\le x\le a$. We need the standard integral from Chapter 7 — Integrals (Section 7.3): $$\int\sqrt{a^2-x^2}\,dx=\dfrac x2\sqrt{a^2-x^2}+\dfrac{a^2}{2}\sin^{-1}\dfrac xa+C.$$

📐 Theorem 8.5 — Area of a circle

For $a\gt0$, the region enclosed by the circle $x^2+y^2=a^2$ has area $\pi a^2$. Equivalently, $\displaystyle\int_0^a\sqrt{a^2-x^2}\,dx=\dfrac{\pi a^2}{4}$.

Proof. By symmetry about both axes, $$A=4\int_0^a\sqrt{a^2-x^2}\,dx=4\Big[\dfrac x2\sqrt{a^2-x^2}+\dfrac{a^2}2\sin^{-1}\dfrac xa\Big]_0^a=4\Big[\Big(0+\dfrac{a^2}2\cdot\dfrac\pi2\Big)-(0+0)\Big]=\pi a^2,$$ using the principal values $\sin^{-1}1=\tfrac\pi2$ and $\sin^{-1}0=0$. $\blacksquare$

✏️ Example 4 — A segment of a circle

Find the area of the smaller region (the minor segment) cut off from the circle $x^2+y^2=16$ by the line $x=2$.

💡 Show solution▶

Sketch Radius $4$. The line $x=2$ meets the circle at $(2,\pm2\sqrt3)$. The smaller region lies to the right of the line, for $2\le x\le4$, and is symmetric about the $x$-axis.

Set up $A=2\displaystyle\int_2^4\sqrt{16-x^2}\,dx$. With $F(x)=\dfrac x2\sqrt{16-x^2}+8\sin^{-1}\dfrac x4$: $F(4)=0+8\cdot\dfrac\pi2=4\pi$ and $F(2)=1\cdot\sqrt{12}+8\cdot\dfrac\pi6=2\sqrt3+\dfrac{4\pi}3$.

Answer $A=2\Big(4\pi-2\sqrt3-\dfrac{4\pi}3\Big)$ $=\dfrac{16\pi}3-4\sqrt3$ $\approx16.755-6.928=9.83$ square units.

Check Geometry: the chord subtends an angle $\tfrac{2\pi}3$ at the centre (since $\cos\theta=\tfrac24$ gives $\theta=\tfrac\pi3$ on each side). Sector $=\tfrac12\cdot16\cdot\tfrac{2\pi}3=\tfrac{16\pi}3$, minus the triangle with vertices $(0,0)$, $(2,\pm2\sqrt3)$, of area $\tfrac12\cdot4\sqrt3\cdot2=4\sqrt3$ ✓.

✏️ Example 5 — Part of a quarter-circle

Find the area of the region in the first quadrant bounded by the circle $x^2+y^2=4$, the $x$-axis and the lines $x=0$ and $x=1$.

💡 Show solution▶

Set up In the first quadrant the circle is $y=\sqrt{4-x^2}\ge0$, so $A=\displaystyle\int_0^1\sqrt{4-x^2}\,dx$.

Integrate $$A=\Big[\dfrac x2\sqrt{4-x^2}+2\sin^{-1}\dfrac x2\Big]_0^1=\dfrac{\sqrt3}2+2\cdot\dfrac\pi6=\dfrac{\sqrt3}2+\dfrac\pi3\approx1.913\ \text{square units}.$$

Check The region lies between the rectangles $0\le x\le1$, $0\le y\le\sqrt3$ (area $1.732$) and $0\le x\le1$, $0\le y\le2$ (area $2$) ✓. Geometrically it is a triangle of area $\tfrac{\sqrt3}2$ plus a sector of angle $\tfrac\pi6$ and area $\tfrac12\cdot4\cdot\tfrac\pi6=\tfrac\pi3$ — the two terms of the answer.

Ellipses: Why the Area Is $\pi ab$

The ellipse $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ ($a,b\gt0$) is also symmetric about both axes. Solving for $y$ in the first quadrant gives $y=\dfrac ba\sqrt{a^2-x^2}$: every vertical strip of the ellipse is the corresponding strip of the circle $x^2+y^2=a^2$ multiplied by $\dfrac ba$. So the ellipse is the circle squashed (or stretched) vertically by the factor $\dfrac ba$ — and its area should be $\pi a^2\cdot\dfrac ba=\pi ab$. Interactive 2 lets you watch this happen; the theorem makes it exact.

📐 Theorem 8.6 — Area of an ellipse

For $a,b\gt0$, the region enclosed by the ellipse $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ has area $\pi ab$. (For $b=a$ this is the circle's $\pi a^2$.)

Proof. By symmetry about both axes, and using Theorem 8.5 for the last integral, $$A=4\int_0^a\dfrac ba\sqrt{a^2-x^2}\,dx=\dfrac{4b}{a}\int_0^a\sqrt{a^2-x^2}\,dx=\dfrac{4b}{a}\cdot\dfrac{\pi a^2}{4}=\pi ab.\ \blacksquare$$

The result does not depend on which axis is longer: for $\dfrac{x^2}{b^2}+\dfrac{y^2}{a^2}=1$ the area is again $\pi ab$. Horizontal strips give the same answer.

✏️ Example 6 — Whole ellipses and a segment

(a) Find the area enclosed by the ellipse $4x^2+9y^2=36$. (b) Find the area of the part of the ellipse $\dfrac{x^2}{16}+\dfrac{y^2}{9}=1$ that lies to the right of the line $x=2$.

💡 Show solution▶

(a) Divide by $36$: $\dfrac{x^2}{9}+\dfrac{y^2}{4}=1$, so $a=3$, $b=2$ and $A=\pi\cdot3\cdot2=6\pi\approx18.85$ square units. (Not $36\pi$: put the equation in standard form first.)

(b) set up Here $a=4$, $b=3$ and the upper half is $y=\tfrac34\sqrt{16-x^2}$. The region with $2\le x\le4$ is symmetric about the $x$-axis: $A=2\cdot\dfrac34\displaystyle\int_2^4\sqrt{16-x^2}\,dx$.

(b) integrate From Example 4, $\displaystyle\int_2^4\sqrt{16-x^2}\,dx=\dfrac{8\pi}3-2\sqrt3$. So $A=\dfrac32\Big(\dfrac{8\pi}3-2\sqrt3\Big)$ $=4\pi-3\sqrt3$ $\approx12.566-5.196=7.37$ square units.

Check The ellipse is the circle of Example 4 squashed by $\tfrac34$, so the segment's area should be $\tfrac34$ of the circle's: $\tfrac34\Big(\tfrac{16\pi}3-4\sqrt3\Big)=4\pi-3\sqrt3$ ✓.

🧭 Problem-solving strategy — areas of conic regions

Standard form first. Rewrite the equation as $y^2=4ax$, $x^2=4ay$, $x^2+y^2=a^2$ or $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ and read off $a$ (and $b$).
Find the corners: where the bounding line meets the curve and the axes.
Use symmetry. Parabolas in standard form are symmetric about their axis; circles and ellipses about both axes. Integrate over the first-quadrant part and multiply by $2$ or $4$ — but only if the region really is symmetric.
Choose the strip so that one formula describes the whole strip: $y=2\sqrt{ax}$ for $y^2=4ax$ (vertical), $x=2\sqrt{ay}$ for $x^2=4ay$ (horizontal).
Integrate using $\displaystyle\int\sqrt{a^2-x^2}\,dx$ for circles and ellipses, and evaluate $\sin^{-1}$ at principal values.
Check with a known result: $\tfrac23$ of the bounding rectangle for a parabolic segment; $\pi a^2$, $\pi ab$ for whole curves; sector minus triangle for a circular segment.

⚠️ Common mistakes when studying alone

1. Forgetting the lower half. The region bounded by $y^2=4ax$ and $x=c$ lies on both sides of the $x$-axis. $\displaystyle\int_0^c2\sqrt{ax}\,dx$ is only half of it.

2. Multiplying by $4$ when the region is only symmetric about one axis. A segment $x\ge c$ of a circle is symmetric about the $x$-axis only: multiply by $2$, not $4$.

3. Reading $a$ and $b$ from a non-standard equation. $4x^2+9y^2=36$ has $a=3$, $b=2$ (area $6\pi$), not $a=2$, $b=3$ from the coefficients, and certainly not area $36\pi$.

4. Degrees in $\sin^{-1}$. $\sin^{-1}\tfrac12=\tfrac\pi6$, not $30$. An area like $8\cdot30$ instead of $8\cdot\tfrac\pi6$ is a common slip.

5. Using the wrong standard integral. $\displaystyle\int\sqrt{a^2-x^2}\,dx$ has two terms; dropping $\tfrac x2\sqrt{a^2-x^2}$ only works when it vanishes at both limits (as at $0$ and $a$).

6. Confusing $y^2=4ax$ with $x^2=4ay$. The first opens to the right (axis = $x$-axis), the second upwards (axis = $y$-axis). Sketch before integrating.

🚀 Beyond the board syllabus (JEE Main · JEE Advanced)

This box is not examined in the Board exam — the area between two curves has been removed from the Board syllabus, but JEE Main and JEE Advanced ask it often. If $f$ and $g$ are continuous and $f(x)\ge g(x)$ on $[a,b]$, a vertical strip runs from the lower curve to the upper one, so its length is $f(x)-g(x)$ and $$A=\int_a^b\big[f(x)-g(x)\big]\,dx\qquad(\text{upper minus lower}).$$ The limits are usually the $x$-coordinates of the points where the curves meet.

Worked example. Find the area enclosed by the parabola $y=x^2$ and the line $y=x+2$.

They meet where $x^2=x+2$, i.e. $(x-2)(x+1)=0$: $x=-1$ and $x=2$. Between them the line is above the parabola (at $x=0$: $2\gt0$). So $$A=\int_{-1}^2(x+2-x^2)\,dx=\Big[\dfrac{x^2}2+2x-\dfrac{x^3}3\Big]_{-1}^2=\dfrac{10}3-\Big(-\dfrac76\Big)=\dfrac92\ \text{square units}.$$ (A useful JEE check: the area between $y=x^2$ and any line meeting it at $x=\alpha,\beta$ is $\tfrac{|\beta-\alpha|^3}{6}=\tfrac{27}6=\tfrac92$ ✓.)

Exercises

Regions bounded by straight lines. Use integration, then check with geometry.

1 The triangle bounded by the line $x+y=4$ and the coordinate axes.
2 The region bounded by $y=x-1$, the $x$-axis and the lines $x=2$ and $x=5$.
3 The region bounded by $y=2x-4$, the $x$-axis and the lines $x=0$ and $x=3$.
4 The triangle bounded by $3x+4y=12$ and the coordinate axes, using horizontal strips.
5 The region bounded by $y=|x-2|$, the $x$-axis and the lines $x=0$ and $x=5$.
6 Find $m\gt0$ if the region bounded by $y=mx$, the $x$-axis and the line $x=3$ has area $18$ square units.

💡 Show solutions▶

1 $\displaystyle\int_0^4(4-x)\,dx=\Big[4x-\tfrac{x^2}2\Big]_0^4=8$ square units $=\tfrac12\cdot4\cdot4$ ✓.

2 $x-1\gt0$ on $(2,5)$: $\Big[\tfrac{x^2}2-x\Big]_2^5=\tfrac{15}2-0=\tfrac{15}2$ square units (a trapezium with sides $1$ and $4$, width $3$ ✓).

3 Root $x=2$. $\displaystyle\int_0^2(2x-4)\,dx=\Big[x^2-4x\Big]_0^2=-4$ and $\displaystyle\int_2^3(2x-4)\,dx=-3-(-4)=1$. Area $=4+1=5$ square units.

4 $x=\tfrac{12-4y}3$ for $0\le y\le3$: $\displaystyle\int_0^3\tfrac{12-4y}{3}\,dy$ $=\tfrac13\Big[12y-2y^2\Big]_0^3$ $=\tfrac13(36-18)=6$ square units $=\tfrac12\cdot4\cdot3$ ✓.

5 Split at $2$: $\displaystyle\int_0^2(2-x)\,dx+\int_2^5(x-2)\,dx=2+\tfrac92=\tfrac{13}2$ square units.

6 $\displaystyle\int_0^3mx\,dx=\tfrac92m=18$, so $m=4$.

Parabolas.

1 The region bounded by $y^2=8x$ and its latus rectum.
2 The region in the first quadrant bounded by $y^2=x$, the $x$-axis and the lines $x=1$ and $x=4$.
3 The region bounded by $y^2=4x$ and the line $x=3$.
4 The region in the first quadrant bounded by $x^2=4y$, the $y$-axis and the lines $y=2$ and $y=4$.
5 The region bounded by $x^2=9y$ and the line $y=4$.
6 The region bounded by $y=x^2$ and the line $y=9$.
7 The line $x=c$ divides the region bounded by $y^2=4x$ and $x=4$ into two parts of equal area. Find $c$.
8 Find $a\gt0$ if the region bounded by $y^2=4ax$ and its latus rectum has area $24$ square units.

💡 Show solutions▶

1 $a=2$, latus rectum $x=2$: $2\displaystyle\int_0^2\sqrt{8x}\,dx=2\cdot2\sqrt2\cdot\tfrac23\cdot2\sqrt2=\tfrac{32}3$ square units $=\tfrac83a^2$ ✓.

2 $y=\sqrt x$: $\tfrac23\Big[x^{3/2}\Big]_1^4=\tfrac23(8-1)=\tfrac{14}3$ square units.

3 $2\displaystyle\int_0^3 2\sqrt x\,dx=4\cdot\tfrac23\cdot3\sqrt3=8\sqrt3\approx13.86$ square units (Theorem 8.4 with $a=1$, $c=3$ ✓).

4 Horizontal strips, $x=2\sqrt y$: $\displaystyle\int_2^4 2\sqrt y\,dy=\tfrac43\big(8-2\sqrt2\big)=\tfrac{32-8\sqrt2}{3}\approx6.90$ square units.

5 $x=\pm3\sqrt y$; by symmetry $A=2\displaystyle\int_0^4 3\sqrt y\,dy=6\cdot\tfrac23\cdot8=32$ square units. Check: the bounding rectangle is $12\times4=48$, and $\tfrac23\cdot48=32$ ✓.

6 $x=\pm\sqrt y$; $A=2\displaystyle\int_0^9\sqrt y\,dy=\tfrac43\cdot27=36$ square units ($\tfrac23$ of the $6\times9$ rectangle ✓).

7 The area to the left of $x=c$ is $\tfrac83c^{3/2}$ and the whole region has area $\tfrac83\cdot8=\tfrac{64}3$. So $\tfrac83c^{3/2}=\tfrac{32}3$, $c^{3/2}=4$ and $c=4^{2/3}=2\sqrt[3]2\approx2.52$.

8 $\tfrac83a^2=24$, so $a^2=9$ and $a=3$.

III

Circles and ellipses.

1 The area enclosed by $x^2+y^2=25$, using integration.
2 The area of the part of $x^2+y^2=9$ in the first quadrant.
3 The region in the first quadrant bounded by $x^2+y^2=16$, the $x$-axis and the lines $x=0$ and $x=2$.
4 The smaller region cut off from $x^2+y^2=4$ by the line $x=1$.
5 The area enclosed by $\dfrac{x^2}{25}+\dfrac{y^2}{16}=1$.
6 The area enclosed by $9x^2+4y^2=36$.
7 The region in the first quadrant bounded by $\dfrac{x^2}{9}+\dfrac{y^2}{4}=1$, the $x$-axis and the lines $x=0$ and $x=\tfrac32$.
8 An ellipse has semi-major axis $5$ and the same area as a circle of radius $4$. Find its semi-minor axis.

💡 Show solutions▶

1 $4\displaystyle\int_0^5\sqrt{25-x^2}\,dx$ $=4\Big[\tfrac x2\sqrt{25-x^2}+\tfrac{25}2\sin^{-1}\tfrac x5\Big]_0^5$ $=4\cdot\tfrac{25}2\cdot\tfrac\pi2=25\pi$ square units.

2 $\displaystyle\int_0^3\sqrt{9-x^2}\,dx=\tfrac{9\pi}4\approx7.07$ square units.

3 $\Big[\tfrac x2\sqrt{16-x^2}+8\sin^{-1}\tfrac x4\Big]_0^2$ $=\sqrt{12}+8\cdot\tfrac\pi6$ $=2\sqrt3+\tfrac{4\pi}3\approx7.65$ square units.

4 $2\displaystyle\int_1^2\sqrt{4-x^2}\,dx$ with $F(x)=\tfrac x2\sqrt{4-x^2}+2\sin^{-1}\tfrac x2$: $F(2)=\pi$, $F(1)=\tfrac{\sqrt3}2+\tfrac\pi3$. $A=2\Big(\tfrac{2\pi}3-\tfrac{\sqrt3}2\Big)=\tfrac{4\pi}3-\sqrt3\approx2.46$ square units.

5 $a=5$, $b=4$: $20\pi\approx62.83$ square units.

6 $\dfrac{x^2}4+\dfrac{y^2}9=1$: semi-axes $2$ and $3$, area $6\pi$ square units (the major axis is vertical; the formula is the same).

7 $y=\tfrac23\sqrt{9-x^2}$: $A=\tfrac23\Big[\tfrac x2\sqrt{9-x^2}+\tfrac92\sin^{-1}\tfrac x3\Big]_0^{3/2}$ $=\tfrac23\Big(\tfrac34\cdot\tfrac{3\sqrt3}2+\tfrac92\cdot\tfrac\pi6\Big)$ $=\tfrac{3\sqrt3}4+\tfrac\pi2\approx2.87$ square units.

8 $\pi\cdot5\cdot b=\pi\cdot16$, so $b=3.2$.

Applications.

1 Elliptical lawn. A lawn has the shape $\dfrac{x^2}{400}+\dfrac{y^2}{100}=1$ (metres). Find its area, and the cost of turfing it at $\text{₹}\,80$ per m².
2 Parabolic window. A window is the region under the arch $y=3-\tfrac34x^2$ above its sill $y=0$ (metres). Find the area of glass needed.
3 Satellite orbit. A satellite's orbit is an ellipse with semi-axes $7000$ km and $6990$ km, and one orbit takes $98$ minutes. By Kepler's law of equal areas, how much area does the line from the Earth to the satellite sweep out per minute?
4 Pond and path. A circular pond $x^2+y^2=100$ (metres) is crossed by a straight footbridge along the line $x=6$. Find the area of the smaller part of the pond.
5 Archimedes' rule. For $k\gt0$, show that the region bounded by $y=k-x^2$ and the $x$-axis has exactly $\tfrac23$ of the area of its bounding rectangle.

💡 Show solutions▶

1 $a=20$, $b=10$: area $200\pi\approx628.3\ \text{m}^2$; cost $\approx628.32\times80\approx\text{₹}\,50\,265$.

2 The arch meets the sill at $x=\pm2$. $\displaystyle\int_{-2}^2\Big(3-\tfrac34x^2\Big)dx$ $=2\Big[3x-\tfrac{x^3}4\Big]_0^2$ $=2(6-2)=8\ \text{m}^2$ ($\tfrac23$ of the $4\times3$ rectangle ✓).

3 Area $=\pi\cdot7000\cdot6990\approx1.537\times10^8\ \text{km}^2$; per minute $\approx\dfrac{1.537\times10^8}{98}\approx1.57\times10^6\ \text{km}^2$.

4 $2\displaystyle\int_6^{10}\sqrt{100-x^2}\,dx$ with $F(x)=\tfrac x2\sqrt{100-x^2}+50\sin^{-1}\tfrac x{10}$: $F(10)=25\pi$, $F(6)=24+50\sin^{-1}\tfrac35$. $A=50\pi-48-100\sin^{-1}\tfrac35\approx44.73\ \text{m}^2$.

5 It meets the axis at $x=\pm\sqrt k$. $A=\displaystyle\int_{-\sqrt k}^{\sqrt k}(k-x^2)\,dx$ $=2\Big(k\sqrt k-\tfrac{k\sqrt k}3\Big)$ $=\tfrac43k^{3/2}$. The rectangle has width $2\sqrt k$ and height $k$: area $2k^{3/2}$. Ratio $\tfrac43\div2=\tfrac23$ ✓.

🎯 Exam corner Board · 1 markBoard · 2 marksBoard · 3 marksBoard · 5 marksJEE MainJEE Advanced

How this is examined: the Board's 3- and 5-mark area questions are almost always one of the regions in this section — a parabola with a vertical line, a circle or a part of it, or an ellipse — and reward a neat sketch, symmetry, the integral $\int\sqrt{a^2-x^2}\,dx$ and correct principal values. JEE adds the area between two curves.

E1. Board · 1 mark The area enclosed by the ellipse $\dfrac{x^2}{16}+\dfrac{y^2}{25}=1$ is (A) $20\pi$ (B) $41\pi$ (C) $80\pi$ (D) $9\pi$

💡 Show solution▶

Idea Semi-axes $4$ and $5$; area $\pi ab$.

Answer (A) $20\pi$. Trap (C) uses $4ab\pi$ (multiplying by $4$ for symmetry twice); (B) adds $16+25$.

E2. Board · 2 marks Find the area of the region bounded by the parabola $y^2=16x$ and its latus rectum.

💡 Show solution▶

Set up $4a=16$, so $a=4$ and the latus rectum is $x=4$. By symmetry $A=2\displaystyle\int_0^4 4\sqrt x\,dx$. (1 mark)

Answer $A=8\cdot\tfrac23\cdot8=\tfrac{128}3\approx42.67$ square units $\big(=\tfrac83a^2\big)$. (1 mark)

E3. Board · 3 marks Find the area of the region in the first quadrant bounded by the circle $x^2+y^2=36$, the $x$-axis and the lines $x=0$ and $x=3$.

💡 Show solution▶

Set up $y=\sqrt{36-x^2}$ in the first quadrant; $A=\displaystyle\int_0^3\sqrt{36-x^2}\,dx$. (1 mark)

Integrate $\Big[\tfrac x2\sqrt{36-x^2}+18\sin^{-1}\tfrac x6\Big]_0^3$ $=\tfrac32\sqrt{27}+18\cdot\tfrac\pi6$. (1 mark)

Answer $A=\tfrac{9\sqrt3}2+3\pi\approx7.79+9.42=17.22$ square units. (1 mark)

E4. Board · 5 marks Using integration, find the area enclosed by the ellipse $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$. Hence find the area enclosed by $4x^2+25y^2=100$ and the area of the part of it that lies in the first quadrant.

💡 Show solution▶

Sketch, symmetry The ellipse is symmetric about both axes; in the first quadrant $y=\tfrac ba\sqrt{a^2-x^2}$, $0\le x\le a$. So $A=4\cdot\tfrac ba\displaystyle\int_0^a\sqrt{a^2-x^2}\,dx$. (1 mark)

Integrate $\displaystyle\int_0^a\sqrt{a^2-x^2}\,dx$ $=\Big[\tfrac x2\sqrt{a^2-x^2}+\tfrac{a^2}2\sin^{-1}\tfrac xa\Big]_0^a$ $=\tfrac{a^2}2\cdot\tfrac\pi2=\tfrac{\pi a^2}4$. (2 marks)

Result $A=\tfrac{4b}a\cdot\tfrac{\pi a^2}4=\pi ab$. (1 mark)

Apply $4x^2+25y^2=100$ gives $\tfrac{x^2}{25}+\tfrac{y^2}4=1$: $a=5$, $b=2$, area $10\pi\approx31.42$ square units; first quadrant $\tfrac{5\pi}2\approx7.85$ square units. (1 mark)

E5. JEE Main The area of the region $\{(x,y):y^2\le4x,\ 0\le x\le1\}$ is (A) $\tfrac43$ (B) $\tfrac83$ (C) $\tfrac{16}3$ (D) $2$

💡 Show solution▶

Method $y^2\le4x$ means $-2\sqrt x\le y\le2\sqrt x$: both halves. $A=2\displaystyle\int_0^12\sqrt x\,dx=\tfrac83$.

Answer (B). Trap (A) counts only $y\ge0$. ⏱ Fast route: this is the parabola $y^2=4x$ ($a=1$) with its latus rectum, so $\tfrac83a^2=\tfrac83$.

E6. JEE Advanced (Numerical answer.) Let $A$ be the area of the region enclosed by the parabola $y=x^2$ and the line $y=2x+3$. Find the value of $3A$.

💡 Show solution▶

Meet $x^2=2x+3$ gives $(x-3)(x+1)=0$: $x=-1$ and $x=3$; between them the line is on top (at $x=0$: $3\gt0$).

Integrate $A=\displaystyle\int_{-1}^3(2x+3-x^2)\,dx$ $=\Big[x^2+3x-\tfrac{x^3}3\Big]_{-1}^3$ $=9-\Big(-\tfrac53\Big)=\tfrac{32}3$ (see the 🚀 box).

Answer $3A=\mathbf{32}$. ⏱ Fast route: $\tfrac{|\beta-\alpha|^3}6=\tfrac{4^3}6=\tfrac{32}3$.

🔬 Research–Teaching NexusMini project · about 60 minutes

Research question. Many cricket grounds and athletics infields are roughly oval. How well does the ellipse formula $\pi ab$ predict the playing area of a real ground?

Why it matters. Groundstaff need the playing area to order grass seed, fertiliser and irrigation water; event planners need it for crowd capacity; and cricket boards set minimum boundary sizes. Surveyors and town planners routinely approximate curved plots by standard shapes and then check the error.

What to do.

Open a free online map with a distance-measuring tool and find an oval cricket ground (or your school field). Measure the longest boundary-to-boundary distance $2a$ and the perpendicular width $2b$ through the centre.
Compute $\pi ab$. Then measure the area directly with the map's area tool (click points round the boundary) — this is your reference value.
Model the boundary as $\tfrac{x^2}{a^2}+\tfrac{y^2}{b^2}=1$. Measure the actual half-width of the ground at five or six positions along the long axis and compare with $\tfrac ba\sqrt{a^2-x^2}$. Where does the ground bulge beyond the ellipse, and where does it fall short?
Estimate the area with vertical strips from your measured half-widths (a spreadsheet sum $2\sum y\,\Delta x$), and compare all three values.

Tools: An online map with distance and area tools, spreadsheet, calculator.

Hand in: A screenshot of your measurements, the values of $a$, $b$ and $\pi ab$, the map-tool area, the strip estimate, the percentage differences, and two sentences on whether "an ellipse" is a fair model of your ground.

Going further. Some grounds are closer to a "stadium" shape (a rectangle with two semicircles). Find a formula for its area and decide which model fits your ground better.

Instructor note: for a ground with axes $150$ m and $130$ m, $a=75$, $b=65$ and $\pi ab\approx15\,315\ \text{m}^2$. Map-tool areas of real oval grounds are usually within $3$–$6\%$ of $\pi ab$; grounds with straighter sides come out larger than the ellipse.

🧪 Self-check quiz

0/5 correct

Q1. The area enclosed by the circle $x^2+y^2=9$ is

Q2. The area enclosed by the ellipse $\dfrac{x^2}{9}+\dfrac{y^2}{4}=1$ is

Q3. The area of the region bounded by the parabola $y^2=4ax$ and its latus rectum is

Q4. $\displaystyle\int_0^a\sqrt{a^2-x^2}\,dx$ (with $a\gt0$) equals

Q5. To find the area of the part of the circle $x^2+y^2=16$ to the right of $x=2$, the correct set-up is

← 8.1 Area under a Curve Chapter 9 — Differential Equations →