Areas, surface areas, volumes via integration

At a Glance

• Frequency: 4 sub-parts across 4 of 13 years (2016, 2017, 2018, 2021)
• Priority tier: T3
• Marks (count): 13 (1), 15 (3)
• Average solve time: ~8 min
• Difficulty mix: medium 2, easy 2
• Section: A | Dominant type: computation

Why This Chapter Matters

Every appearance of this atom has been a 13–15 mark question in Section A — high reward for moderate effort. The four archetypes cover the complete UPSC repertoire: surface area of a tilted plane (2016), volume under a paraboloid cap (2017), volume of an elliptic solid of revolution (2018), and area of an astroid via Beta functions (2021). The two most common errors are forgetting the Jacobian when changing coordinates for an ellipse, and computing the wrong special case of the Beta function. Get these right and the marks follow mechanically.

Minimum Theory

Surface area of a graph. For $z = f(x,y)$ over a region $R$: $S = \iint_R \sqrt{1 + z_x^2 + z_y^2}\,dA.$ When $z = z(x,y)$ is a plane, $z_x$ and $z_y$ are constants and the radicand is a constant — the surface area is just (constant factor) × (projected area).

Volume by a double integral. Volume of the solid between $z = z_1(x,y)$ and $z = z_2(x,y)$ over region $R$: $V = \iint_R \big(z_2(x,y) - z_1(x,y)\big)\,dA.$

Volume of revolution — disc method. Revolving $y = f(x) \ge 0$ around the $x$-axis: at each station $x$, the cross-section is a disc of radius $f(x)$: $V = \pi\int_a^b [f(x)]^2\,dx.$

Parametric area. For a closed curve $x = x(t)$, $y = y(t)$ traversed counter-clockwise: $A = \oint x\,dy = \int x(t)\,\dot y(t)\,dt.$

Beta function. $\int_0^{\pi/2}\cos^m\theta\sin^n\theta\,d\theta = \dfrac{1}{2}B\!\left(\dfrac{m+1}{2},\dfrac{n+1}{2}\right)$, where $B(p,q) = \dfrac{\Gamma(p)\Gamma(q)}{\Gamma(p+q)}$. Key values: $\Gamma(1/2) = \sqrt\pi$, $\Gamma(3/2) = \tfrac{1}{2}\sqrt\pi$, $\Gamma(5/2) = \tfrac{3}{4}\sqrt\pi$.

Question Archetypes

surface-area (1 question; 2016)

Recognition Cues

• “Find the surface area of the plane … cut off by the cylinder … and the coordinate planes.”
• The surface is a plane (so $z_x, z_y$ are constants and the integrand is constant).

Solution Template

1. Express the surface as $z = z(x,y)$ from the plane equation. Compute $z_x, z_y$.
2. Compute the area factor $\sqrt{1 + z_x^2 + z_y^2}$. For a plane this is a constant.
3. Identify the projected region $R$ in the $xy$-plane: the projection of the given boundaries (usually a disc or sector).
4. Surface area = (area factor) × (area of $R$).

Worked Example

2016 Paper 1, 2016-P1-Q3c (15 marks)

Find the surface area of $x + 2y + 2z = 12$ cut off by $x = 0$, $y = 0$, and $x^2 + y^2 = 16$.

From $z = (12 - x - 2y)/2$: $z_x = -1/2$, $z_y = -1$.

Area factor: $\sqrt{1 + 1/4 + 1} = \sqrt{9/4} = 3/2$.

Projected region $R$: $x \ge 0$, $y \ge 0$, $x^2 + y^2 \le 16$ — the quarter disc of radius 4. Area of $R$ = $\tfrac{1}{4}\pi(4)^2 = 4\pi$.

$\boxed{S = \frac{3}{2} \times 4\pi = 6\pi.}$

Common Traps

• The cylinder $x^2+y^2=16$ plus $x=0$, $y=0$ bounds a quarter disc (first quadrant only), not the full disc. Area = $4\pi$, not $16\pi$.
• The plane $x+2y+2z=12$ has normal $(1,2,2)$, $|(1,2,2)| = 3$, so the area scaling factor is $3/2$ (the $z$-component of the unit normal is $2/3$, and $1/(2/3) = 3/2$). Both routes give the same answer.

volume-by-integration (1 question; 2017)

Recognition Cues

• “Find the volume of the solid above the $xy$-plane and below the surface $z = f(x,y)$ cut off by $z = c$.”
• The base region is often an ellipse — use stretched polar coordinates $x = ar\cos\theta$, $y = br\sin\theta$.

Solution Template

1. Identify the base region $R$: where the surface meets the cap $z = c$, i.e., $f(x,y) = c$.
2. Set up the volume: $V = \iint_R [c - f(x,y)]\,dA$.
3. Change to stretched polar coordinates if $R$ is an ellipse: $x = ar\cos\theta$, $y = br\sin\theta$, Jacobian $= abr$.
4. Evaluate the resulting iterated integral.

Worked Example

2017 Paper 1, 2017-P1-Q2a (15 marks)

Volume above the $xy$-plane, below $x^2 + y^2/4 = z$, cut off by $z = 9$.

Base region: $x^2 + y^2/4 \le 9$ (ellipse, semi-axes $3$ and $6$). Height: $9 - (x^2 + y^2/4)$.

Stretched polar: $x = r\cos\theta$, $y = 2r\sin\theta$, so $x^2 + y^2/4 = r^2$, Jacobian $= 2r$.

$V = \int_0^{2\pi}\int_0^3 (9 - r^2) \cdot 2r\,dr\,d\theta = 2\pi \int_0^3 (18r - 2r^3)\,dr = 2\pi\left[9r^2 - \frac{r^4}{2}\right]_0^3 = 2\pi \cdot \frac{81}{2} = 81\pi.$

$\boxed{V = 81\pi.}$

Common Traps

• The Jacobian for $x = r\cos\theta$, $y = 2r\sin\theta$ is $2r$ (not $r$). Missing the factor of 2 halves the answer.
• The semi-axes of the ellipse $x^2 + y^2/4 = 9$ are $a=3$ and $b=6$ (not $b=3$).
• The integrand is the gap $9 - (x^2 + y^2/4)$, not the paraboloid value itself.

volume-of-revolution (1 question; 2018)

Recognition Cues

• “The ellipse $x^2/a^2 + y^2/b^2 = 1$ revolves about the $x$-axis. Find the volume.”
• Disc method: $V = \pi\int y^2\,dx$.

Solution Template

1. Express $y^2$ from the curve equation: $y^2 = b^2(1 - x^2/a^2)$.
2. Disc formula: $V = \pi\int_{-a}^{a} y^2\,dx = 2\pi b^2\int_0^a (1 - x^2/a^2)\,dx$.
3. Evaluate the standard integral; sanity-check via the $a=b$ sphere case.

Worked Example

2018 Paper 1, 2018-P1-Q2c (13 marks)

Volume of solid formed by revolving $x^2/a^2 + y^2/b^2 = 1$ about the $x$-axis.

$V = \pi\int_{-a}^a b^2\!\left(1-\frac{x^2}{a^2}\right)dx = 2\pi b^2\left[x - \frac{x^3}{3a^2}\right]_0^a = 2\pi b^2 \cdot \frac{2a}{3} = \frac{4}{3}\pi a b^2.$

$\boxed{V = \frac{4}{3}\pi ab^2.}$

Sanity check: $a = b = r$ gives $\tfrac{4}{3}\pi r^3$ (sphere volume) ✓.

Common Traps

• Revolution about the $x$-axis means radius = $y$, integrate in $x$ → volume $= \tfrac{4}{3}\pi ab^2$. About the $y$-axis: radius = $x$, integrate in $y$ → $\tfrac{4}{3}\pi a^2 b$. The asymmetry in $a$ vs $b$ catches many students.
• Use symmetry ($0$ to $a$ and double) to avoid sign errors.

area-by-integration (1 question; 2021)

Recognition Cues

• “Show the area of the astroid $x^{2/3} + y^{2/3} = a^{2/3}$ is $3\pi a^2/8$.”
• Parametric form $x = a\cos^3\theta$, $y = a\sin^3\theta$ combined with the Beta function.

Solution Template

1. Parametrize: use $x = a\cos^3\theta$, $y = a\sin^3\theta$.
2. Area formula: $A = \oint x\,dy = \int_0^{2\pi} x(\theta)\dot y(\theta)\,d\theta$.
3. Compute $\dot y$: $\dot y = 3a\sin^2\theta\cos\theta$.
4. Integral becomes: $A = 3a^2\int_0^{2\pi}\cos^4\theta\sin^2\theta\,d\theta = 3a^2\cdot 4\int_0^{\pi/2}\cos^4\theta\sin^2\theta\,d\theta$ (by symmetry).
5. Apply Beta formula: $\int_0^{\pi/2}\cos^4\theta\sin^2\theta\,d\theta = \tfrac{1}{2}B(5/2, 3/2) = \tfrac{1}{2}\cdot\dfrac{\Gamma(5/2)\Gamma(3/2)}{\Gamma(4)} = \dfrac{\pi}{32}$.
6. Combine: $A = 3a^2\cdot 4\cdot\dfrac{\pi}{32} = \dfrac{3\pi a^2}{8}$.

Worked Example

2021 Paper 1, 2021-P1-Q4b (15 marks)

Show the area of the astroid $x^{2/3}+y^{2/3}=a^{2/3}$ is $\dfrac{3\pi a^2}{8}$.

Following the template: $B(5/2, 3/2) = \dfrac{\Gamma(5/2)\Gamma(3/2)}{\Gamma(4)} = \dfrac{(3/4)\sqrt\pi \cdot (1/2)\sqrt\pi}{6} = \dfrac{(3/8)\pi}{6} = \dfrac{\pi}{16}$.

So $\int_0^{\pi/2}\cos^4\theta\sin^2\theta\,d\theta = \dfrac{\pi}{32}$, and $A = 3a^2 \times 4 \times \dfrac{\pi}{32} = \dfrac{3\pi a^2}{8}$. $\square$

Common Traps

• The four-fold symmetry gives $\int_0^{2\pi}(\cdots) = 4\int_0^{\pi/2}(\cdots)$. Don’t integrate $0$ to $\pi/2$ and forget the factor of 4.
• $\Gamma(5/2) = \tfrac{3}{2}\cdot\tfrac{1}{2}\cdot\sqrt\pi = \tfrac{3}{4}\sqrt\pi$ (recursive formula $\Gamma(n+1)=n\Gamma(n)$, not $\Gamma(n+1)=n!$).
• The Beta formula gives $\int_0^{\pi/2}\cos^m\sin^n = \tfrac{1}{2}B((m+1)/2,(n+1)/2)$ — note the $\tfrac{1}{2}$ factor.

Marks-Aware Writing

13–15 mark computation: Write the setup equation (area/volume integral) on one line, the substitution on the next, the intermediate integral, and the final evaluation. Show the substitution Jacobian and the Beta function computation explicitly — the examiner is checking each step. Conclude with a sanity check (e.g. the sphere limit for the ellipsoid).

Practice Set

• 2024-P1-Q3b (20 m) — volume computation;
• 2023-P1-Q2b (15 m) — area/volume;
• 2022-P1-Q3b (15 m) — similar structure;
• 2018-P1-Q1d (10 m) — short area computation;