Question

Evaluate ∫∫∫z^2 dxdydz taken over the volume bounded by the surfaces x^2+y^2=a^2, x^2+y^2=z and z=0.

(a) \(\pi \frac{a^8}{12} \)

(b) \(\pi \frac{a^8}{4} \)

(c) \(\pi \frac{a^4}{4} \)

(d) \(\pi \frac{a^8}{8} \)

I got this question by my school teacher while I was bunking the class.

My question comes from Volume Integrals in chapter Vector Integral Calculus of Engineering Mathematics

Answer 1

Correct answer is (a) \(\pi \frac{a^8}{12} \)

The explanation: z: 0 to \(x^2+y^2 \)

\( y: \pi \frac{a^8}{12} \) to \( \sqrt{a^2-x^2} \)

x: -a to a

\(\displaystyle\int\int\int z^2 dxdydz = \int_{-a}^a \int_{-\sqrt{(a^2-x^2 )}}^{\sqrt{a^2-x^2}}\int_0^{x^2+y^2}z^2 dxdydz \)

\(\displaystyle = \int_{-a}^a \int_{-\sqrt{a^2-x^2}}^{\sqrt{a^2-x^2}}\frac{z^3}{3} dxdy \) from 0 to \( x^2+y^2. \)

\(\displaystyle = \int_{-a}^a \int_{-\sqrt{a^2-x^2}}^{\sqrt{a^2-x^2}} \frac{(x^2+y^2 )^3}{3} dxdy\)

\( = \frac{1}{3} \int_0^π \int_{-a}^a r^7 drdθ \)

\( = π \frac{a^8}{12}. \)