Find the volume V of the solid bounded by the cylinder $x^2 +y^2 = 1$, the xy-plane and the plane $x + z = 1 $.

Question

Find the volume V of the solid bounded by the cylinder $x^2 +y^2 = 1$, the xy-plane and the plane $x + z = 1 $.

Hi all, i cant seem to get the correct answer for this question. The answer is $\pi$ but i got $2\pi$ . Was hoping that someone could check to see what i'm doing wrong. Thanks in advance.

I tried doing this with polar coordinates. so $x^2+y^2=r^2$ , $x=rcos\theta$ , $y=rsin\theta$

$0\le \theta \le 2\pi$ , $0 \le r \le 1 $ and $ 0 \le z \le 1 - r cos\theta$

Did the integration like this

$\int_0^{2\pi}\int_0^1\int_0^{1-rcos\theta} dz dr d\theta$

= $\int_0^{2\pi}\int_0^1 (1-rcos\theta) dr d\theta$

= $\int_0^{2\pi} (1- \frac{cos\theta}{2}) d\theta$

=$2\pi$

edit: mistake was that r is missing eg.$\int_0^{2\pi}\int_0^1\int_0^{1-rcos\theta} r dz dr d\theta$

Answer 1

Project the whole situation onto the $(x,z)$-plane, and draw a figure showing this plane. You will then realize that the body $B$ in question is half of the cylinder $$\bigl\{(x,y,z)\>|\>x^2+y^2\leq 1, \ 0\leq z\leq2\bigr\}\ .$$ It follows that ${\rm vol}(B)={1\over2}\cdot \pi\cdot 2=\pi$.

Answer 2

Saw off the portion above$z=1$ and flip it around to fit with the rest of the solid, making a cylinder of radius $1$ and height $1$. Thus $V=\pi r^2h=\pi$.