# VOLUME INTEGRATION HELP PLZ!

• Dec 9th 2008, 10:15 PM
coogle0925
VOLUME INTEGRATION HELP PLZ!
Q: Find the volume of the solid formed by
z=x^2 +y^2 and z=2y-4x+1

i think im supposed to do cylindrical coordinates.
i got the limits for z and theta
but not for r
my limits were
z: r^2 => 2rsin(theta) - 4rcos(theta) +1
Theta: 0 => 2pi

CAN U HELP ME PLZ? THANK YOU
• Dec 10th 2008, 11:36 AM
shawsend
I think r is a function of $\displaystyle \theta$ and the solution of:

$\displaystyle (x+1)^2+(y-1)^2=6$

when $\displaystyle x=r\cos(\theta),\; y=r\sin(\theta)$

That's the domain in the x-y plane where the integration is being done (just complete the square when you equate both equations.

Doing that I get (in cylindrical coordinates):

$\displaystyle \int_0^{2\pi}\int_0^{r(\theta)}\int_{g(r,\theta)}^ {h(r,\theta)} rdzdrd\theta$

where $\displaystyle h(r,\theta)=r^2$

$\displaystyle g(r,\theta)=2r\sin(\theta)-4r\cos(theta)+1$

and $\displaystyle r(\theta)$ is one of the roots to the solution of that equation above. I don't know. Looks pretty messy to me. Maybe another easier way though.