1. ## Find a volume

I'm asked to find the volume which is inside both the sphere $z=x^2+y^2+z^2$ and the paraboloid $z=2(x^2+y^2)$.

My attempt: The equation of the sphere : $z=x^2+y^2+z^2 \Leftrightarrow \frac{1}{4}=x^2+y^2+(z-\frac{1}{2})$. So it's a sphere with radius $\frac{1}{2}$ and center in $(0,0,\frac{1}{2})$. And the paraboloid is easy to visualize.
$V=\int_0^{2\pi} \int_0^{\pi} \int _?^{?} \rho ^2 \sin (\phi) d\rho d\phi d\theta$.
My guess is $\rho$ goes from the equation of the paraboloid to the equation of the sphere. So from $(\rho \sin (\theta) \cos (\phi))^2+(\rho \sin (\theta) \sin (\phi))^2$ to $(\rho \sin (\theta) \cos (\phi))^2+(\rho \sin (\theta) \sin (\phi))^2+ (\rho \cos(\phi))^2$. I just want to be sure.

2. Just to make sure there's no typo. Is that supposed to be

$z=x^{2}+y^{2}+z^{2}$?.

That z on the left. Is that correct?.

If $x^{2}+y^{2}+z^{2}=2x^{2}+2y^{2}$, then $x^{2}+y^{2}=z^{2}$

$r^{2}=z^{2}$

3. Originally Posted by galactus
Just to make sure there's no typo. Is that supposed to be

$z=x^{2}+y^{2}+z^{2}$?.

That z on the left. Is that correct?.
Yes it is correct. (No typo error)

If , then

Ok, so $r$ (my $\rho$) goes from $0$ to $(\rho \cos (\phi))^2$?

4. Originally Posted by arbolis
I'm asked to find the volume which is inside both the sphere $z=x^2+y^2+z^2$ and the paraboloid $z=2(x^2+y^2)$.

My attempt: The equation of the sphere : $z=x^2+y^2+z^2 \Leftrightarrow \frac{1}{4}=x^2+y^2+(z-\frac{1}{2})$. So it's a sphere with radius $\frac{1}{2}$ and center in $(0,0,\frac{1}{2})$. And the paraboloid is easy to visualize.
$V=\int_0^{2\pi} \int_0^{\pi} \int _?^{?} \rho ^2 \sin (\phi) d\rho d\phi d\theta$.
My guess is $\rho$ goes from the equation of the paraboloid to the equation of the sphere. So from $(\rho \sin (\theta) \cos (\phi))^2+(\rho \sin (\theta) \sin (\phi))^2$ to $(\rho \sin (\theta) \cos (\phi))^2+(\rho \sin (\theta) \sin (\phi))^2+ (\rho \cos(\phi))^2$. I just want to be sure.
the boundaries of $\rho$ it is wrong you have

$z=x^2+y^2+z^2 \Rightarrow \frac{1}{4}=x^2+y^2+(z-\frac{1}{2})^2$

$z=2(x^2+y^2)$ you know that

$x=\rho cos\theta sin\phi , y=\rho sin\theta sin\phi , z=\rho cos\phi$

$\rho$ change from the parabola to the sphere as you said
we need to present $\rho$ with respect to $\phi,\theta$

let z'=z-1/2 I just move the solid 1/2 unit down the new equations

$\frac{1}{4}=x^2+y^2+z^2$

$\frac{1}{4}=\rho ^2$

$\rho=\frac{1}{2}$ since you need the positive upper coordinate ...........(1)

$z+\frac{1}{2}=2(x^2+y^2)$

$\rho \cos\phi+\frac{1}{2} =2(\cos^2\theta \sin^2\phi \rho^2 +\sin^2\theta \sin^2\phi \rho^2)$

$\rho \cos\phi +\frac{1}{2}=2(sin^2\phi \rho^2)$

$2\rho^2 \sin^2\phi -\rho \cos\phi -\frac{1}{2}=0$

$\rho = \frac{cos\phi \mp \sqrt{\cos^2\phi -4(2sin^2\phi)(\frac{-1}{2})}}{4sin^2\phi}$

$\rho = \frac{cos\phi \mp \sqrt{\cos^2\phi +4(sin^2\phi)}}{4sin^2\phi}$

$\rho = \frac{cos\phi \mp \sqrt{1+3(sin^2\phi)}}{4sin^2\phi}$ take

$\rho = \frac{cos\phi + \sqrt{1+3(sin^2\phi)}}{4sin^2\phi}$ the integral

$\int_0^{2\pi} \int_0^{\pi}\int_{\frac{cos\phi + \sqrt{1+3(sin^2\phi)}}{4sin^2\phi}}^{\frac{1}{2}} d\rho d\phi d\theta$

if you use cylindrical it will be easier I think

5. Originally Posted by Amer

$\int_0^{2\pi} \int_0^{\pi}\int_{\frac{cos\phi + \sqrt{1+3(sin^2\phi)}}{4sin^2\phi}}^{\frac{1}{2}} d\rho d\phi d\theta$

if you use cylindrical it will be easier I think
I'll try it. By the way did you forgot to put the integrand $\rho ^2 \sin \phi$?
I just realized I forgot the ^2 in the expression $\frac{1}{4}=x^2+y^2+(z-\frac{1}{2})$, that was a typo.

6. Originally Posted by arbolis
I'll try it. By the way did you forgot to put the integrand $\rho ^2 \sin \phi$?
I just realized I forgot the ^2 in the expression $\frac{1}{4}=x^2+y^2+(z-\frac{1}{2})$, that was a typo.

ooh I forgot it I concern in the boundaries and forgot the Jacobean ....

7. Originally Posted by arbolis
I'm asked to find the volume which is inside both the sphere $z=x^2+y^2+z^2$ and the paraboloid $z=2(x^2+y^2)$.

My attempt: The equation of the sphere : $z=x^2+y^2+z^2 \Leftrightarrow \frac{1}{4}=x^2+y^2+(z-\frac{1}{2})$. So it's a sphere with radius $\frac{1}{2}$ and center in $(0,0,\frac{1}{2})$. And the paraboloid is easy to visualize.
$V=\int_0^{2\pi} \int_0^{\pi} \int _?^{?} \rho ^2 \sin (\phi) d\rho d\phi d\theta$.
My guess is $\rho$ goes from the equation of the paraboloid to the equation of the sphere. So from $(\rho \sin (\theta) \cos (\phi))^2+(\rho \sin (\theta) \sin (\phi))^2$ to $(\rho \sin (\theta) \cos (\phi))^2+(\rho \sin (\theta) \sin (\phi))^2+ (\rho \cos(\phi))^2$. I just want to be sure.
if you look at the image of your solids in yz plane, you'll have $z=2y^2, \ y^2+(z-1/2)^2=1/4, \ 0 \leq z \leq 1/2.$ see that the parabola is inside the circle in that interval. so the volume inside both

the sphere and the paraboloid is nothing but the volume bounded between $z=2(x^2+y^2)$ and the plane $z=1/2.$ that is: $\int_0^{2 \pi} \int_0^{1/2} (1/2 - 2r^2)r \ dr d \theta=\frac{\pi}{16}.$

8. Originally Posted by NonCommAlg
if you look at the image of your solids in yz plane, you'll have $z=2y^2, \ y^2+(z-1/2)^2=1/4, \ 0 \leq z \leq 1/2.$ see that the parabola is inside the circle in that interval. so the volume inside both

the sphere and the paraboloid is nothing but the volume bounded between $z=2(x^2+y^2)$ and the plane $z=1/2.$ that is: $\int_0^{2 \pi} \int_0^{1/2} (1/2 - 2r^2)r \ dr d \theta=\frac{\pi}{16}.$
I was scared this would happen! I still can't see why the parabola is inside the sphere, that is, why $2y^2 for $0\leq z \leq \frac{1}{2}$. (It doesn't holds if I put $z=2y^2 for $0\leq z \leq \frac{1}{2}$....I've much to understand!) But I'll check it. Thanks by the way.

9. Originally Posted by arbolis
I was scared this would happen! I still can't see why the parabola is inside the sphere, that is, why $2y^2 for $0\leq z \leq \frac{1}{2}$. (It doesn't holds if I put $z=2y^2 for $0\leq z \leq \frac{1}{2}$....I've much to understand!) But I'll check it. Thanks by the way.
to prove that the parabola is inside the circle you need to show that $z_{\text{parabola}}=2y^2 \geq z_{\text{circle}}=1/2 - \sqrt{1/4 - y^2}$ for $|y| \leq 1/2.$ that inequality is equivalent to $1-4y^2 \leq \sqrt{1-4y^2},$ which is

obviously true, because for any $0 \leq \alpha \leq 1$ we have $\alpha \leq \sqrt{\alpha}.$

10. Today a friend of mine told me the right answer is $\frac{7\pi}{48}$. I had shown him your (NCA) post, he copied it in his draft. He went to university and checked the result with a teacher who said it is wrong. My friend didn't understand all what the teacher said, he just came up with $V=\int_0^{2\pi} \int_0^{\frac{1}{2}} \int _{2r^2}^{\frac{1}{2}+\sqrt{\frac{1}{4}-r^2}} r dzdrd\theta = \frac{7\pi}{48}$.
As I trust you and that your post makes sense to me, I wonder where's the error of the teacher. I know he used cylindrical coordinates, but I don't really know what he did. It seems that for him $z$ goes from $2r^2$ (so the paraboloid) to $\frac{1}{2}+\sqrt{\frac{1}{4}-r^2}$, which I don't know what is it (it has no other choice than being the sphere, but as you pointed out, the paraboloid is inside it. I'm quite confused!)
Does someone has anything to say?

11. Originally Posted by arbolis
Today a friend of mine told me the right answer is $\frac{7\pi}{48}$. I had shown him your (NCA) post, he copied it in his draft. He went to university and checked the result with a teacher who said it is wrong. My friend didn't understand all what the teacher said, he just came up with $V=\int_0^{2\pi} \int_0^{\frac{1}{2}} \int _{2r^2}^{\frac{1}{2}+\sqrt{\frac{1}{4}-r^2}} r dzdrd\theta = \frac{7\pi}{48}$.
As I trust you and that your post makes sense to me, I wonder where's the error of the teacher. I know he used cylindrical coordinates, but I don't really know what he did. It seems that for him $z$ goes from $2r^2$ (so the paraboloid) to $\frac{1}{2}+\sqrt{\frac{1}{4}-r^2}$, which I don't know what is it (it has no other choice than being the sphere, but as you pointed out, the paraboloid is inside it. I'm quite confused!)
Does someone has anything to say?
your teacher is right! sorry i forgot to add the volume of the upper half of the sphere, which is $\frac{\pi}{12},$ to $\frac{\pi}{16}.$ so the correct answer is $\frac{\pi}{12}+\frac{\pi}{16}=\frac{7\pi}{48}.$

as i said before the paraboloid is completely inside the sphere for $0 \leq z \leq 1/2.$ so the plane $z=\frac{1}{2}$ divides your solid into two parts: a lower part, which is the paraboloid, and an upper part,

which is the upper half of your sphere. i forgot to add the volume of the upper part of your solid! you should now understand the limits of the integral in your teacher's solution as well.