Triple Integrals; Cylindrical and Spherical Coordinates

At a Glance

• Frequency: 1 sub-part across 1 of 14 years (2023)
• Priority tier: T4
• Marks (count): 15 (1)
• Average solve time: ~15 min
• Difficulty mix: medium 1
• Section: B | Dominant type: computation

Why This Chapter Matters

This atom appeared once in 2023 as a 15-mark Section B computation. UPSC typically asks you to evaluate a triple integral over a geometrically natural region — a sphere, cone, or cylinder — by converting to cylindrical or spherical coordinates. The choice of coordinate system is the key decision: spherical coordinates collapse sphere/cone boundaries to simple constant-limit bounds, while cylindrical coordinates suit cylinders and paraboloids. The computation is mechanical once the limits are set up correctly.

Minimum Theory

Cylindrical Coordinates

$x = r\cos\theta, \quad y = r\sin\theta, \quad z = z, \qquad r \ge 0,\; 0 \le \theta \le 2\pi.$

Volume element: $dV = r\,dr\,d\theta\,dz$.

Useful for: cylinders $x^2 + y^2 = a^2$, paraboloids $z = x^2 + y^2$, cones $z = \sqrt{x^2+y^2}$.

Spherical Coordinates

$x = \rho\sin\phi\cos\theta, \quad y = \rho\sin\phi\sin\theta, \quad z = \rho\cos\phi,$ $\rho \ge 0,\; 0 \le \phi \le \pi,\; 0 \le \theta \le 2\pi.$

Volume element: $dV = \rho^2\sin\phi\,d\rho\,d\phi\,d\theta$.

Note: $\phi$ is the polar angle from the positive $z$-axis (colatitude); $\theta$ is the azimuthal angle in the $xy$-plane.

Also: $x^2 + y^2 + z^2 = \rho^2$.

Useful for: spheres $\rho = a$, cones $\phi = \phi_0$ (constant), hemispheres.

Setting Up Limits

Standard sphere $x^2 + y^2 + z^2 \le a^2$ in spherical: $\rho: 0 \to a$, $\phi: 0 \to \pi$, $\theta: 0 \to 2\pi$.

Upper hemisphere $x^2+y^2+z^2 \le a^2$, $z \ge 0$: $\phi: 0 \to \pi/2$.

Cone below sphere region between $z = \sqrt{x^2+y^2}$ (cone, $\phi = \pi/4$) and $x^2+y^2+z^2 = a^2$ (sphere): $\rho: 0 \to a$, $\phi: 0 \to \pi/4$, $\theta: 0 \to 2\pi$.

Cylinder $x^2+y^2 \le a^2$, $0 \le z \le h$ in cylindrical: $r: 0 \to a$, $\theta: 0 \to 2\pi$, $z: 0 \to h$.

Volume Formulas (from triple integrals)

$V(\text{sphere of radius }a) = \iiint dV = \frac{4}{3}\pi a^3, \qquad V(\text{cone, half-angle }\phi_0, \text{height }h) = \frac{\pi h^3 \tan^2\!\phi_0}{3}.$

Question Archetypes

coordinate-change-evaluation (1 question; 2023)

Recognition Cues

• The region is a sphere, ball, hemisphere, cone, or cylinder
• The integrand involves $x^2+y^2+z^2$, $x^2+y^2$, or is identically 1 (volume computation)
• “Using spherical coordinates” or “using cylindrical coordinates” may be stated, or you must choose
• 15 marks: full setup (change of variables, Jacobian, limit identification) plus iterated integral evaluation expected

Solution Template

1. Identify the region; choose coordinates (spherical if sphere/cone, cylindrical if cylinder/paraboloid).
2. Write the coordinate transformation and Jacobian.
3. Express the region in the new coordinates (find the constant or simple-function bounds for each variable).
4. Rewrite the integrand in the new coordinates.
5. Factor the iterated integral (often the $\theta$ integral separates immediately as $2\pi$).
6. Evaluate inner integrals first, working outward.
7. State the final answer with correct units/form.

Worked Example

2023 Paper 1, 2023-P1-Q5c (15 marks)

Evaluate $\displaystyle\iiint_V (x^2 + y^2 + z^2)\,dV$, where $V$ is the region inside the sphere $x^2 + y^2 + z^2 = a^2$.

Step 1. Choose coordinates.

The region is a full ball of radius $a$ and the integrand is $\rho^2$ in spherical coordinates. Use spherical coordinates.

Step 2. Write transformation and Jacobian.

$x = \rho\sin\phi\cos\theta,\quad y = \rho\sin\phi\sin\theta,\quad z = \rho\cos\phi, \quad dV = \rho^2\sin\phi\,d\rho\,d\phi\,d\theta.$

Integrand: $x^2+y^2+z^2 = \rho^2$.

Step 3. Set up limits.

Full ball: $\rho \in [0, a]$, $\phi \in [0, \pi]$, $\theta \in [0, 2\pi]$.

Step 4. Write iterated integral.

$I = \int_0^{2\pi}\!\!\int_0^{\pi}\!\!\int_0^{a} \rho^2 \cdot \rho^2\sin\phi\,d\rho\,d\phi\,d\theta = \int_0^{2\pi}d\theta \int_0^{\pi}\sin\phi\,d\phi \int_0^{a}\rho^4\,d\rho.$

Step 5. Evaluate each factor.

$\int_0^{2\pi}d\theta = 2\pi.$

$\int_0^{\pi}\sin\phi\,d\phi = \bigl[-\cos\phi\bigr]_0^{\pi} = -\cos\pi + \cos 0 = 1 + 1 = 2.$

$\int_0^{a}\rho^4\,d\rho = \frac{\rho^5}{5}\Bigg|_0^a = \frac{a^5}{5}.$

Step 6. Combine.

$I = 2\pi \cdot 2 \cdot \frac{a^5}{5} = \frac{4\pi a^5}{5}.$

$\boxed{\iiint_V (x^2+y^2+z^2)\,dV = \frac{4\pi a^5}{5}}$

Common Traps

• Forgetting $\sin\phi$ in the Jacobian: The spherical Jacobian is $\rho^2\sin\phi$ — omitting $\sin\phi$ gives the wrong answer and is the single most common error on this topic.
• Confusing $\phi$ and $\theta$ conventions: Some books swap $\phi$ and $\theta$; in the UPSC / standard Indian convention, $\phi$ runs from $0$ to $\pi$ (from $z$-axis) and $\theta$ runs from $0$ to $2\pi$ (in $xy$-plane).
• Cone boundary in spherical coordinates: The cone $z = \sqrt{x^2+y^2}$ corresponds to $\phi = \pi/4$ (not $\phi = \pi/3$ or $\pi/2$); derive it from $\cos\phi = \sin\phi \Rightarrow \tan\phi = 1$.
• Wrong limits for a hemisphere: Upper hemisphere is $z \ge 0$, so $\phi \in [0, \pi/2]$, not $[0, \pi]$.

Marks-Aware Writing

For a 15-mark computation:

• 3 marks — correct choice of coordinate system stated, transformation and Jacobian written explicitly.
• 4 marks — limits for all three variables identified correctly and justified (sketch or verbal description of the region).
• 4 marks — iterated integral written correctly (integrand $\times$ Jacobian, in the right order of integration), factored where possible.
• 4 marks — each integral evaluated correctly, final answer stated in simplified form.

Always write $dV = \rho^2\sin\phi\,d\rho\,d\phi\,d\theta$ (or $r\,dr\,d\theta\,dz$) on the first line of your answer — it signals to the examiner that you know the Jacobian and earns the first mark cluster.

Practice Set

Only one historical question on this atom (shown above).