# Triple Integral question

I need some help in this

The question states find the volume for the region bounded below by z = 3 - 2y and above by z = x^2 + y^2 using cylindrical coordinates .

Now i tried to do it but i got stuck with the part where i find the radius . i found
r^2 = 3-2r*sin(zeta) :( its hard to factorize and use this for the boundary

i already know that z will range from r^2 to 3-2r*sin(zeta) and zeta range from zero to 2 pie

so how can this question be solved ??? :)

btw the correct answer is 8pie

HallsofIvy
Homework Helper
First, you have "below" and "above" reversed. The plane z= 3- 2y is above the paraboloid $z= x^2+ y^2$ in the bounded region.

They intersect where $z= x^2+ y^2= 3- 3y$ which is equivalent to $x^2+ y^2+ 3y= 3$. $x^2+ y^2+ 3y+ 9/4= x^2+ (y+3/2)^2= 3+ 9/4= 21/4$. That is a circle with center at (0, -3/2) which is why ordinary polar coordinates do not give a simple equation. Either
1: integrate with x from $-\sqrt{21}/2$ to $\sqrt{21}/2$ and, for each x, y from $-3/2- \sqrt{21/4- x^2}$ to $-3/2+ \sqrt{21/4- x^2}$.

2: Let $x= rcos(\theta)$ and $y= -3/2+ rsin(\theta)$, shifting the origin to (0, -3/2).

Oh i c :D !!! Thanks alot . I wasn't aware of shifting the origin :) !!

Thanks :D !

LOL 8pie

HallsofIvy