Question

find the volume of the solid bounded by z=-y, the xy plane, and r=4cos(theta)

Answer 1

\[\int\limits_{-\pi/2}^{0}\int\limits_{0}^{4\cos(\theta)}\int\limits_{0}^{-\sin(\theta)}r dz dr d(\theta)\]

Answer 2

my question is: 4cos(theta) is a circle of radius 2 , when i integrate with that in approach \[\int\limits_{-\pi/2}^{0}\int\limits_{0}^{2}\int\limits_{0}^{-\sin(\theta)}r dz dr d(\theta)\]

Answer 3

i am off by a factor of 2, am i doing something wrong!!!

Answer 4

hmmm

Answer 5

Can you draw it?

Answer 6

ill try

Answer 7

No, they are z, r, theta

Answer 8

It's called cylindrical coordinates.

Answer 9

yeah i tried.. no dice

Answer 10

@chris2k59 Did you try to integrate normally?

Answer 11

what do you mean
?

Answer 12

I can't see how integrating over this will work.\[
\int\limits_{-\pi/2}^{0}\int\limits_{0}^{2}\int\limits_{0}^{-\sin(\theta)}r dz dr d(\theta)
\]

Answer 13

So why not integrate over the actual setup?

Answer 14

r=4cos(theta)
r^2=4rcos(theta)
r^2=4x
x^2+y^2=4x
then by completing the square
(x-2)^2+y^2=4; r=2

Answer 15

the original setup works fine, but when used this approach i was off by a factor of 2...
i want to know if i just had a mistake somewhere

Answer 16

is this approach valid???

Answer 17

Unfortunately your approach is not valid because an offset circle is going to have a different domain than a centered circle.

Answer 18

You are changing the domain, which in general fails.

Answer 19

lets make the region first
start with "z"