UPSC 2017 Maths Optional Paper 1 Q2a — Step-by-Step Solution

15 marks · Section A

Areas, surface areas, volumes via integration · Calculus · asked 4× in 13 yrs · Read the full method →

Question

Find the volume of the solid above the $xy$ -plane and directly below the portion of the elliptic paraboloid $x^2+\dfrac{y^2}{4}=z$ which is cut off by the plane $z=9$ .

Technique

Volume $=\iint(\text{top}-\text{bottom})\,dA$ over the elliptical base; the substitution $x=r\cos\theta,\ y=2r\sin\theta$ turns the ellipse into a disc with Jacobian $2r$ .

Solution

Step 1 — Identify the solid and its base

The paraboloid is $z=x^2+\dfrac{y^2}{4}$ , opening upward with vertex at the origin. The plane $z=9$ caps it. The solid is bounded below by $z=0$ (the $xy$ -plane), above by $z=9$ , and on the side by the paraboloid; equivalently it is the region between the paraboloid surface and the cap, projected onto its base.

The base (shadow on the $xy$ -plane) is where the paraboloid meets $z=9$ :

x^2+\frac{y^2}{4}=9,

an ellipse with semi-axes $a=3$ (in $x$ ) and $b=6$ (in $y$ ).

Step 2 — Set up the volume integral

Over the base ellipse the solid has height (cap minus paraboloid):

V=\iint_{x^2+\frac{y^2}{4}\le 9}\Big(9-\big(x^2+\tfrac{y^2}{4}\big)\Big)\,dx\,dy.

Step 3 — Stretched polar coordinates

Let

x=r\cos\theta,\qquad y=2r\sin\theta,\qquad \text{so}\quad x^2+\frac{y^2}{4}=r^2,\quad dx\,dy=2r\,dr\,d\theta.

The base $x^2+\tfrac{y^2}4\le 9$ becomes $0\le r\le 3$ , $0\le\theta\le 2\pi$ . Then

V=\int_{0}^{2\pi}\!\!\int_{0}^{3}(9-r^2)\,2r\,dr\,d\theta.

Step 4 — Evaluate

\int_{0}^{3}(9-r^2)2r\,dr=\int_0^3(18r-2r^3)\,dr=\Big[9r^2-\tfrac{r^4}{2}\Big]_0^3=81-\tfrac{81}{2}=\tfrac{81}{2}.

Multiply by the $\theta$ -range $2\pi$ :

V=2\pi\cdot\frac{81}{2}=81\pi.

Answer

\boxed{\;V=81\pi\ \text{cubic units}.\;}