Question

Hi, need help checking my answer for this question. Find the volume of the wedge-shaped region (Figure 1) contained in the cylinder x^2+y^2=16 and bounded above by the plane z=x and below by the xy-plane.

Answer 1

I converted to cylindrical coordinates:
\[0 \le \theta \le2\pi\]\[0\le r \le 4\]\[0\le z \le x\]

Answer 2

\[-\frac{\pi}{2}\le\theta\le\frac{\pi}{2}\]

right ?

Answer 3

because the wedge is in I and IV octants

Answer 4

not sure if this is correct, but I then set up a triple integral for the volume:
\[\int\limits_{0}^{2\pi}\int\limits_{0}^{4}\int\limits_{0}^{x}r dzdrd \theta\]

Answer 5

ok so it should rather be
\[\int\limits_{-\pi/2}^{\pi/2}\int\limits_{0}^{4}\int\limits_{0}^{x}r dzdrd \theta\] ?

Answer 6

also replace \(x\) by \(r\cos\theta\)

Answer 7

yes, i did that later in my workings

Answer 8

looks good!

Answer 9

would this be correct:
\[\int\limits_{-\pi/2}^{\pi/2}\int\limits_{0}^{4}r^2*\cos(\theta) dr d \theta\]

Answer 10

yes im getting the same

Answer 11

and then? \[\int\limits_{-\pi/2}^{\pi/2}\frac{ 64 }{ 3 }\cos(\theta) d \theta\]

Answer 12

good so far, keep going

Answer 13

\[\frac{ 64 }{ 3 } \int\limits_{-\pi/2}^{\pi/2}\cos(\theta) = (\frac{ 64 }{ 3 }) \sin(\theta) \] with \[\sin(\theta)\] evaluated \[-\pi/2 \to p/2\]

Answer 14

that evaluates to \[\frac{ 64 }{ 3 }(\sin(\pi/2)-\sin(-\pi/2)) = 128/3\] is that correct?

Answer 15

Looks perfect!
http://www.wolframalpha.com/input/?i=%5Cint%5Climits_%7B-pi%2F2%7D%5E%7Bpi%2F2%7D%5Cint%5Climits_%7B0%7D%5E%7B4%7D+%28r%5E2*cos%28%5Ctheta%29%29+dr+d%5Ctheta

Answer 16

cool thanks for you help :)

Answer 17

np:)