Question

use cylindrical coordinates to find the volume of the region in the first octant bounded by cylinder x^2+y^2=4 and the plane z=1-2x+3y

Answer 1

\(\large \iiint \limits_R dz ~r dr d\theta \)

Answer 2

so change to polar, huh..

Answer 3

\(\large \int \limits_0^{\frac{\pi}{2}} ~ \int \limits_0^{2}~~\int \limits_0^{ 1-2r\cos\theta + 3r\sin\theta } dz ~r dr d\theta \)

Answer 4

evaluate

Answer 5

ok, thanks! I just needed help setting up! so it's easier to use polar when you see something like \[x^2 + y^2 \]. but what does "use cylindrical coordinates" mean?

Answer 6

polar = 2D
cylinderical = 3D

Answer 7

thank you for the help!

Answer 8

polar : \(\large (r, \theta)\)
cylindrical : \(\large (r, \theta, z)\)

Answer 9

you may simply setup the double integral for volume, if you're not familiar wid triple integrals...

Answer 10

\( \mathbb{ \int \limits_0^{\frac{\pi}{2}} ~ \int \limits_0^{2}~~\int \limits_0^{ 1-2r\cos\theta + 3r\sin\theta } dz ~r dr d\theta = \int \limits_0^{\frac{\pi}{2}} ~ \int \limits_0^{2}~~ (1-2r\cos\theta + 3r\sin\theta ) ~r dr d\theta }\)

Answer 11

right side is just a double integral

Answer 12

http://www.wolframalpha.com/input/?i=%5Cint_0%5E%28pi%2F2%29+%5Cint_0%5E%282%29+%5Cint_0%5E%7B1-2rcos%28%5Ctheta%29%2B3rsin%28%5Ctheta%29%7D++r++dz+dr+d%5Ctheta

Answer 13

would that give me the same answer? because I am more familiar with double integrals.

Answer 14

yup ! ive just evaluated the inner integral in triple integral :)

Answer 15

\(\mathbb{ \int \limits_0^{\frac{\pi}{2}} ~ \int \limits_0^{2}~~\color{red}{\int \limits_0^{ 1-2r\cos\theta + 3r\sin\theta } dz} ~r dr d\theta = \int \limits_0^{\frac{\pi}{2}} ~ \int \limits_0^{2}~~\color{red}{ (1-2r\cos\theta + 3r\sin\theta )} ~r dr d\theta }\)

Answer 16

the integrand u get when u do double integral is simply the bound for inner integral in triple integral