Calculate the volume of the body defined by $x^2+y^2+z^2≤1$ for which $x^2+z^2≥z$
Attempt:
For the volume, I used this formula: $V=\iint y(x,z)dxdz$
I find it easier to express the problem through $y$ instead of $z$, so I get the following:
$$y(x,z)= \sqrt{(1-x^2-z^2)}$$
$$x^2 + (z - \frac{1}{2 })^2 = ( \frac{1}{2} )^2$$
From these I can get the visual graph of the intersection and visualize the figure that I need to calculate the volume from.
Next I switch to polar coordinates:
$$x = r\cos(\phi)$$ $$z = r\sin(\phi)$$ $$|J| = r$$
From this, I get the integral bounds: $0 ≤ r ≤ \sin(\phi)$ and $0 ≤ \phi ≤ \pi$
For the first integral I get the following: $$\int \sqrt{1-r^2}rdr$$ with the respected bounds.
After the first integration, I get the following: $$ \frac{1}{3} \int (1-\cos^6(\phi))d\phi$$ with the respected bounds.
The final result is: $V = \dfrac{11\pi}{48}$.
Can anyone confirm if this is the correct result?
The problem asks you to compute the volume of the region inside the sphere $x^2+y^2+z^2=1$ and outside the cylinder $x^2+(z-1/2)^2=(1/2)^2$ whose boundary cannot be expressed in the form $y=f(x,z)$ for some $f$.
The volume of the region that is shared by both the cylinder and the sphere can be expressed as an iterated triple integral:
$$\int_0^\pi \int_0^{\sin(\theta)} \int_{-\sqrt{1-r^2}}^{\sqrt{1-r^2}}\mathrm{dy}r\mathrm{d}r\mathrm{d}\theta$$ Now subtract this from $4\pi/3$ to get your answer.