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Math Help - Find a volume

  1. #1
    MHF Contributor arbolis's Avatar
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    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.
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  2. #2
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    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}
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  3. #3
    MHF Contributor arbolis's Avatar
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    Quote Originally Posted by galactus View Post
    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?
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  4. #4
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by arbolis View Post
    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
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  5. #5
    MHF Contributor arbolis's Avatar
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    Quote Originally Posted by Amer View Post

    \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.
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  6. #6
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by arbolis View Post
    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 ....
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  7. #7
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    Quote Originally Posted by arbolis View Post
    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}.
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  8. #8
    MHF Contributor arbolis's Avatar
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    Quote Originally Posted by NonCommAlg View Post
    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<y^2+(z-1/2)^2=1/4 for 0\leq z \leq \frac{1}{2}. (It doesn't holds if I put z=2y^2<y^2+(z-1/2)^2=1/4 for 0\leq z \leq \frac{1}{2}....I've much to understand!) But I'll check it. Thanks by the way.
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    Quote Originally Posted by arbolis View Post
    I was scared this would happen! I still can't see why the parabola is inside the sphere, that is, why 2y^2<y^2+(z-1/2)^2=1/4 for 0\leq z \leq \frac{1}{2}. (It doesn't holds if I put z=2y^2<y^2+(z-1/2)^2=1/4 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}.
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  10. #10
    MHF Contributor arbolis's Avatar
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    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?
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  11. #11
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    Quote Originally Posted by arbolis View Post
    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.
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