Answer to Question #179753 in Calculus for Sourav

Question

Evaluate the integral of f(x, y, z) = x + z - 3 over the cylinder bounded by x^2 + y^2 = 1, z = 0 and z = 1.

Answer 1

the integral of f(x,y,z)=x+z-3 over the cylinder $x^2+y^2=1$ ; z=0 and z=1 is;

$\int^{1}_{0}\int^{1}_{-1}\int^{\sqrt{1-y^2}} _{-\sqrt{1-y^2}}(x+z-3)dxdydz$

$=\int^{1}_{0}\int^{1}_{-1}[(\frac{x^2}{2}+xz-3x)]^{\sqrt{1-y^2}}_{-\sqrt{1-y^2}}dydz$

$=\int^{1}_{0}\int^{1}_{-1}2(z-3)\sqrt{1-y^2} dydz$

$=\int^{1}_{0}[2(z-3)\int^{1}_{-1}\sqrt{1-y^2}]dydz$

by letting $y=sin \theta,\implies dy=cos\theta d\theta$ we then may evaluate the integral as

$=\int^{1}_{0}[2(z-3)\int^{1}_{-1}\sqrt{1-y^2}dy]dz$

$=\int^{1}_{0}[2(z-3)\int^{\pi/2}_{-\pi/2}\sqrt{1-sin ^2 \theta} cos\theta d\theta ]dz$

$=\int^{1}_{0}[2(z-3)\int^{\pi/2}_{-\pi/2}cos ^2 \theta d\theta ]dz$

$=\int^{1}_{0}[2(z-3). \frac{1}{2}\int^{\pi/2}_{-\pi/2}(1+cos2\theta) d\theta ]dz$

$=\int^{1}_{0}(z-3)[\theta+\frac{1}{2}sin 2\theta]^{\pi/2}_{-\pi/2}dz$

$=\int^{1}_{0}(z-3)[(\frac{\pi}{2}+\frac {1}{2}sin \pi)- (-\frac{\pi}{2}+\frac{1}{2}sin \pi]dz$

$=\pi\int^{1}_{0}(z-3)dz$

$=\pi[\frac{z^2}{2}-3z]^{1}_{0}$

$=\pi(\frac{-5}{2})= \frac{-5}{2}\pi$