# Multiple Integration Volume

• Aug 1st 2006, 09:11 AM
MikeWakefield32
Multiple Integration Volume
Compute the volume of the region under the plane z=2x+3y+30 and over the region in the xy plane bounded by the circle (x^2)+(y^2)=2y.

I tried using polar coordinates, but I have no idea what the appropriate range of theta would be. I'm also confused with the 2y as part of the circle equation. Any help with the answer would be appreciated.
• Aug 1st 2006, 09:45 AM
CaptainBlack
Change coordinates in the x-y plane to $x'=x, y'=y-1$, then
in terms of the new coordinate the region of integration is
$x'^2+y'^2=1$,
and the plane is now $z=2x'+3y'+33$.

Now switch the cylindrical polars, so $x'=r\cos \theta$, $y'=r\sin \theta$.
Then
the region of integration is,
$0\leq r\leq 1$, $0\leq \theta < 2\pi$

RonL
• Aug 1st 2006, 10:43 AM
ThePerfectHacker
Quote:

Originally Posted by MikeWakefield32
Compute the volume of the region under the plane z=2x+3y+30 and over the region in the xy plane bounded by the circle (x^2)+(y^2)=2y.

I tried using polar coordinates, but I have no idea what the appropriate range of theta would be. I'm also confused with the 2y as part of the circle equation. Any help with the answer would be appreciated.

Let me explain diffrently.

The function,
$z=2x+3y+30$
Can be expressed as, using polar substitution.
$z=2r\cos \theta +3r \sin \theta +30$
Futhermore, the region,
$x^2+y^2=2y$
Is,
$r^2=2 r\sin \theta$
Thus,
$r=2\sin \theta$

Therefore, the double integral is, (remember to multiply by "r"),
$\int_0^{\pi} \int_0^{2\sin \theta}2r^2\cos \theta +3r^2\sin \theta +30r\, dr\, d\theta$