Question

a little confused....trying to find a volume using double integrals and getting an answer of zero which is obviously wrong.....

Find the volume of the wedge-shaped region (Figure 1) contained in the cylinder x^2+y^2=36 and bounded above by the plane z=x and below by the xy-plane.

Answer 1

so i did \[\int\limits_{0}^{2\pi}\int\limits_{0}^{6}rcos(\theta) r drd(\theta)\]
\[\int\limits_{0}^{6}r^2\cos(\theta)dr =72\cos(\theta).\]
\[\int\limits_{0}^{2\pi}72\cos(\theta) d(\theta)=0\]

Answer 2

you are using polar coordinates

Answer 3

yeah, this section is on polar coordinates.

Answer 4

@ganeshie8, are you able to help me again?

Answer 5

@zepdrix or someone else? question due in an hour. thanks!

Answer 6

Polar coordinates won't be much help. Use cylindrical. (Okay, cylindrical is made up of polar, but there's another dimension involved.)

Answer 7

The space filled by the wedge is described by the set of points,
\[D:=\{(x,y,z)~:~0\le x\le6,~-\sqrt{36-x^2}\le y\le\sqrt{36-x^2},~0\le z\le x\}\]
(note: not the only way to describe the region)

Transforming to polar coordinates, you have
\[D:=\left\{(r,\theta,z)~:~0\le r\le6,~-\frac{\pi}{2}\le \theta\le\frac{\pi}{2},~0\le z\le r\cos\theta\right\}\]
and the volume would be given by
\[\int\int\int_DdV=\int_{-\pi/2}^{\pi/2}\int_0^6\int_0^{6r\cos\theta} r~dz~dr~d\theta\]

Answer 8

oh. umm. we haven't done this yet. triple integrals is next section. that's awesome they threw this in here since i don't know how to do it yet....

Answer 9

Well you can still technically reduce this to a double integral in polar. It's the same as
\[\int_{-\pi/2}^{\pi/2}\int_0^6r(6r\cos\theta)~dr~d\theta\]
Ignore my comment about the uselessness of polar coordinates in this situation. I'm more used to finding volume with triple integrals.

Answer 10

why did the 2pi change to pi/2?

Answer 11

When \(\theta=0\), you have a ray on the positive \(x\) axis. The wedge has its \(\theta\) ranging from the fourth to the first quadrant.

Answer 12

ok. lets see if i can work that out now

Answer 13

ok, i got it using my original setup but changing my theta bounds. still r^2costheta. answer was 144.

Answer 14

Your answer, or expected answer?

Answer 15

both. Entered Answer Preview Result
144
144
correct

Answer 16

Hmm, I guess you don't need that 6 in there... strange.

Answer 17

i think the dr bound covers the 6 radius in there. thanks for the help!

Answer 18

Oh my mistake, I tacked the extra 6 for no reason haha. The limits in the set notation for \(D\) are correct.

Answer 19

And you're welcome!