Question

Find the volume of the cylinder bounded by x^2+y^2 = 4, y+z = 4 and z=0.

(a) \(16π-\frac{32}{3} \)

(b) \(32π-\frac{32}{3} \)

(c) \(16-32 \frac{π}{3} \)

(d) \(32-32 \frac{π}{3} \)

This question was posed to me in a national level competition.

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

Answer 1

Right option is (a) \(16π-\frac{32}{3} \)

The best I can explain: \(z: (4-y)\)

\(x: -\sqrt{4-y^2} \) to \( \sqrt{4-y^2} \)

y: -2 to 2

Volume = \(\displaystyle\int_{-2}^2 \int_{-\sqrt{4-y^2}}^{\sqrt{4-y^2}} (4-y)dxdy \)

\(= 2\int_{-2}^2 (4y-y^2 )dxdy \) from 0 to \( \sqrt{4-y^2} \)

\(= 4\big\{4(0 + π) – 4 + \frac{4}{3}\big\} \)

\(= 16π-\frac{32}{3}. \)