how solve this 3D integral

$\displaystyle \int_{0}^{1}\int_{x}^{\sqrt{x}}\int_{0}^{\sqrt{x-y^{2}}}dzdydx$

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

$\displaystyle z=\sqrt{x-y^{2}}$ -> $\displaystyle z^2+y^2=x$

so we have a circle

but the radius changes from x to $\displaystyle \sqrt{x}$.

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

$\displaystyle \int_{0}^{1}\int_{0}^{2\pi}\int_{x}^{\sqrt{x}}rdrd \theta dz$

but i dont know how mathmatickly to change the intervals?

how to solve it in a polar way?

Re: how solve this 3D integral

Can you provide the actual and entire problem statement?

Re: how solve this 3D integral

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

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.

$\displaystyle y \in [0,1]$

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

Re: how solve this 3D integral

how it whould make it easier

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.

$\displaystyle y \in [0,1]$

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

it worked

how to solve

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

Re: how solve this 3D integral

Complete the square: $\displaystyle y- y^2= -(y^2- y+ 1/4- 1/4)= -(y- 1/2)^2+ 1/4$. Let u= y- 1/2 and that becomes $\displaystyle 1/4- u^2$ so the further substitution v= (1/2)sin(t) gives $\displaystyle (1/4)- (1/4)sin^2(t)= (1/4)(1- sin^2t)= (1/4)cos^2(t)$. That makes $\displaystyle (y- y^2)^{3/2}= ((1/4)cos^2(t))^{3/2}= (1/8)cos^3(t)$.