# how solve this 3D integral

• June 30th 2011, 01:46 PM
transgalactic
how solve this 3D integral
$\int_{0}^{1}\int_{x}^{\sqrt{x}}\int_{0}^{\sqrt{x-y^{2}}}dzdydx$

i tried to traw the projection of the top interval in the z-y plane

$z=\sqrt{x-y^{2}}$ -> $z^2+y^2=x$
so we have a circle
but the radius changes from x to $\sqrt{x}$.

which is not possible because when we scan the ring it goes from the biiger radius to the smaller one.
then i thought to try in anyway ,and i wrote the integral in polar
$\int_{0}^{1}\int_{0}^{2\pi}\int_{x}^{\sqrt{x}}rdrd \theta dz$

but i dont know how mathmatickly to change the intervals?
how to solve it in a polar way?
• June 30th 2011, 02:31 PM
TKHunny
Re: how solve this 3D integral
Can you provide the actual and entire problem statement?
• June 30th 2011, 02:47 PM
transgalactic
Re: how solve this 3D integral
find the volume enclosed by x=y z=0 and $y^2+z^2=x$
• June 30th 2011, 07:05 PM
TKHunny
Re: how solve this 3D integral
I'm not seeing the problem. It looks like you have it. Of course, your life may be way easier if you switch x and y.

$y \in [0,1]$

$x \in \left[y^{2},y\right]$
• July 1st 2011, 12:23 AM
transgalactic
Re: how solve this 3D integral
how it whould make it easier
• July 2nd 2011, 03:29 AM
transgalactic
Re: how solve this 3D integral
Quote:

Originally Posted by TKHunny
I'm not seeing the problem. It looks like you have it. Of course, your life may be way easier if you switch x and y.

$y \in [0,1]$

$x \in \left[y^{2},y\right]$

it worked
how to solve
$\int (y-y^2)^{\frac{3}{2}}$
• July 2nd 2011, 09:05 AM
HallsofIvy
Re: how solve this 3D integral
Complete the square: $y- y^2= -(y^2- y+ 1/4- 1/4)= -(y- 1/2)^2+ 1/4$. Let u= y- 1/2 and that becomes $1/4- u^2$ so the further substitution v= (1/2)sin(t) gives $(1/4)- (1/4)sin^2(t)= (1/4)(1- sin^2t)= (1/4)cos^2(t)$. That makes $(y- y^2)^{3/2}= ((1/4)cos^2(t))^{3/2}= (1/8)cos^3(t)$.