UPSC 2023 Maths Optional Paper 1 Q2b — Step-by-Step Solution

15 marks · Section A

Triple Integrals; Cylindrical and Spherical Coordinates · Calculus · Read the full method →

Question

Evaluate the triple integral which gives the volume of the solid enclosed between the two paraboloids $Z=5(x^2+y^2)$ and $Z=6-7x^2-y^2$ .

Technique

Find intersection projection; integrate the difference of the two surfaces over that region. A linear scaling $u=x\sqrt 2,\,v=y$ converts the elliptic region to the unit disk, then polar coordinates.

Solution

Step 1 — Find the curve of intersection.

Set $5(x^2+y^2)=6-7x^2-y^2$ :

5x^2+5y^2=6-7x^2-y^2\Rightarrow 12x^2+6y^2=6\Rightarrow 2x^2+y^2=1.

Projection onto $xy$ -plane is the ellipse $D=\{(x,y):2x^2+y^2\le 1\}$ .

$Axial cross-section (y=0) of the solid between the two paraboloids. The lower paraboloid Z=5(x^2+y^2) (blue) opens upward from the vertex O; the upper paraboloid Z=6-7x^2-y^2 (red) opens downward from (0,0,6). They meet on the circle whose cross-section is the pair of points (\pm 1/\sqrt2,\,5/2); the shaded lens is the region whose volume is sought — the upper surface lies above the lower throughout the projection 2x^2+y^2\le 1.$

Step 2 — Volume integral.

The upper paraboloid $z=6-7x^2-y^2$ is above the lower $z=5(x^2+y^2)$ on $D$ . Volume:

V=\iint_D\bigl[(6-7x^2-y^2)-5(x^2+y^2)\bigr]\,dA=\iint_D[6-12x^2-6y^2]\,dA.

Step 3 — Change of variables to circular coordinates.

Let $u=x\sqrt 2,\,v=y$ (so $2x^2+y^2=u^2+v^2$ ). Jacobian: $|\partial(x,y)/\partial(u,v)|=1/\sqrt 2$ . Then $D$ corresponds to $u^2+v^2\le 1$ .

Substitute $x=u/\sqrt 2,\,y=v$ :

6-12x^2-6y^2=6-12(u^2/2)-6v^2=6-6u^2-6v^2=6(1-u^2-v^2).

V=\iint_{u^2+v^2\le 1}6(1-u^2-v^2)\cdot\frac{1}{\sqrt 2}\,du\,dv=\frac{6}{\sqrt 2}\iint_{u^2+v^2\le 1}(1-u^2-v^2)\,du\,dv.

Step 4 — Polar coordinates.

Set $u=r\cos\theta,\,v=r\sin\theta$ :

\iint_{u^2+v^2\le 1}(1-u^2-v^2)\,du\,dv=\int_0^{2\pi}\!\!\int_0^1(1-r^2)\,r\,dr\,d\theta=2\pi\cdot\Bigl[\frac{r^2}{2}-\frac{r^4}{4}\Bigr]_0^1=2\pi\cdot\frac{1}{4}=\frac{\pi}{2}.

Therefore

V=\frac{6}{\sqrt 2}\cdot\frac{\pi}{2}=\frac{3\pi}{\sqrt 2}=\frac{3\pi\sqrt 2}{2}.

Answer

\boxed{\;V=\frac{3\pi\sqrt 2}{2}=\frac{3\pi}{\sqrt 2}.\;}