Question

The domain is the right circular cylinder whose base is the circle r=3cos(theta) and whose top lies in the plane z=5-x

Answer 1

that could be 0 to pi/2 for theta...

Answer 2

@zepdrix

Answer 3

@some_someone

Answer 4

@some_someone it's calculus iv

Answer 5

it's conversions from Cartesian to cylinder coordinates. I know the dz and the rdr but the dtheta is a pain

Answer 6

http://www.geom.uiuc.edu/docs/reference/CRC-formulas/node42.html

Answer 7

?????????? wut?! I'm aware of the formulas...just finding the theta which is a circle that looks like a cr2032 battery when it comes to cylindrical.

Answer 8

@tkhunny

Answer 9

Looks like you have it except the limits on \(\theta\). Try \(\pi/2 \le \theta \le \pi/2\). You do not want to tract the whole thing twice.

Answer 10

how did you get -pi/2 to pi/2 for the theta?

Answer 11

because well I know for 0 to 2pi the circle is really huge...like....

Answer 12

I tend to want to start at a logical spot and return to the same spot, if I can. The Origin in the x-y Plan seemed logical. Where is the \(\cos(\theta) = 0\)?

Answer 13

cos theta is 0 when it's pi/2

Answer 14

0 to 2pi starts out as far from the origin as possible and traces the entire path TWICE. That is definitely too much.

Answer 15

Right, or -pi/2. There are other choices to tract the whole thing ONCE. That's just the one I picked.

Answer 16

traces the path...from -pi/2 to pi/2... if it was 0 to 2pi that circle would've been way off the cylinder

Answer 17

is there any easier way to get the theta without feeling confused/.

Answer 18

?? No, \(\theta\) has nothing to do with size are the radius of the area. That's 'r'. You can stop feeling confused when you stop confusing \(\theta\) with r. \(\theta\) just points. It does not say how far.

Answer 19

theta is a point in the circle?

Answer 20

No, neither r not \(\theta\) is sufficient to determine a point. It takes both. \(\theta\) says which direction and r says how far from the Origin.

You have symmetry about the x-axis. You could do \(0 \le \theta \le \pi/2\), and multiply the whole thing by 2. That might be less confusing.

Answer 21

that would be 0 < theta < pi if I multiply twice

Answer 22

No, do NOT play with the limits. 0 to pi/2 is half the area. Multiply the result by 2.

Like I said earlier, there are many choices for covering the entire area. \(0 \le \theta \le \pi\) will be fine. Just cover the entire circle and don't do it twice (like 0 to 2pi does.)

Answer 23

You want to pick limits that make your life better, not worse. Getting a zero in there is usually helpful. Symmetry is often beneficial.