# Thread: how solve this 3D integral

1. ## 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?

2. ## Re: how solve this 3D integral

Can you provide the actual and entire problem statement?

3. ## Re: how solve this 3D integral

find the volume enclosed by x=y z=0 and $y^2+z^2=x$

4. ## 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.

$y \in [0,1]$

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

5. ## Re: how solve this 3D integral

how it whould make it easier

6. ## Re: how solve this 3D integral

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
$\int (y-y^2)^{\frac{3}{2}}$

7. ## 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 $(y- y^2)^{3/2}= ((1/4)cos^2(t))^{3/2}= (1/8)cos^3(t)$.