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Math Help - integration, area above cone and under sphere

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    integration, area above cone and under sphere

    Given a cone z=\sqrt{x^2+y^2} and a sphere x^2+y^2+z^2=1, find the area above the cone and under the sphere.

    I figure if I convert this into polar coordinates it will simplify the calculations, so I'll have 0 \leq r \leq 1 for the radius since x^2+y^2+z^2=1. Now for the angles, I'm not 100% sure, but I would imagine that they're in a \frac{\pi}{4} neighborhood on the xz and yz planes, but if we consider the xy plane then this will just be a circle centered at 0 with radius 1.

    I figure I'll have:

    \iint_D \sqrt{x^2+y^2} dA = \int \int_0^1 \sqrt{r^2} \cdot r \ dr \ d\theta = \int \int_0^1 r^2 \cdot  \ dr \ d\theta but without the values for \theta, I can't continue.
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    Quote Originally Posted by lllll View Post
    Given a cone z=\sqrt{x^2+y^2} and a sphere x^2+y^2+z^2=1, find the area above the cone and under the sphere.

    I figure if I convert this into polar coordinates it will simplify the calculations, so I'll have 0 \leq r \leq 1 for the radius since x^2+y^2+z^2=1. Now for the angles, I'm not 100% sure, but I would imagine that they're in a \frac{\pi}{4} neighborhood on the xz and yz planes, but if we consider the xy plane then this will just be a circle centered at 0 with radius 1.

    I figure I'll have:

    \iint_D \sqrt{x^2+y^2} dA = \int \int_0^1 \sqrt{r^2} \cdot r \ dr \ d\theta = \int \int_0^1 r^2 \cdot \ dr \ d\theta but without the values for \theta, I can't continue.
    I have a question. Is it area you're looking for or volume? The reason I ask is because it's worded weird. If you replace the word area with volume it make more sense.
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    Assuming it's volume, let's try spherical coordinates?.

    The equation of the sphere is {\rho}=1

    the equation of the cone is z=\sqrt{x^{2}+y^{2}}

    {\rho}cos({\phi})=\sqrt{{\rho}^{2}sin^{2}({\phi})c  os^{2}({\theta})+{\rho}^{2}sin^{2}({\phi})sin^{2}(  {\theta})}

    which simplifies down to:

    {\rho}cos({\phi})={\rho}sin({\rho})

    tan({\phi})=1

    Therefore, {\phi}=\frac{\pi}{4} and we have:

    \int_{0}^{2\pi}\int_{0}^{\frac{\pi}{4}}\int_{0}^{1  }{\rho}^{2}sin({\phi})d{\rho}d{\phi}d{\theta}
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    Quote Originally Posted by danny arrigo View Post
    I have a question. Is it area you're looking for or volume? The reason I ask is because it's worded weird. If you replace the word area with volume it make more sense.
    It's supposed to be volume.

    Quote Originally Posted by galactus View Post
    Assuming it's volume, let's try spherical coordinates?.
    I haven gotten to to spherical coordinates yet, is there a way of solving this using polar coordinates.

    doing a little more work on this problem, the closest I got was

    \int_0^{2\pi} \int_{1/\sqrt{2}}^1 r^2 \ dr \ d\theta = \frac{1}{6} \pi (4-\sqrt{2})

    after substituting z=\sqrt{x^2+y^2} into x^2+y^2+z^2=1 and setting z=0.

    although this solution is incorrect it's supposed to be \frac{2}{3} \pi (\sqrt{2}-1)
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    Quote Originally Posted by lllll View Post
    It's supposed to be volume.



    I haven gotten to to spherical coordinates yet, is there a way of solving this using polar coordinates.

    doing a little more work on this problem, the closest I got was

    \int_0^{2\pi} \int_{1/\sqrt{2}}^1 r^2 \ dr \ d\theta = \frac{1}{6} \pi (4-\sqrt{2})

    after substituting z=\sqrt{x^2+y^2} into x^2+y^2+z^2=1 and setting z=0.

    although this solution is incorrect it's supposed to be \frac{2}{3} \pi (\sqrt{2}-1)
    For volume you need to solve

    \iint\limits_R f(x,y)-g(x,y) dA

    Here, f and g are your surfaces

    f(x,y) = \sqrt{1-x^2-y^2}\;\;g(x,y)=\sqrt{x^2+y^2} (I'm assuming the top half of the sphere)

    For the region of integration, we find the intersection

    \sqrt{1-x^2-y^2}=\sqrt{x^2+y^2}

    or

    x^2+y^2 = \frac{1}{2} (so the region is a circle)

    polar coordinates is definitely the way to go

    \int_0^{2 \pi} \int_0^{\frac{1}{\sqrt{2}}}\left( \sqrt{1-r^2} - r\right) r dr d \theta\; = \; \frac{2 \pi}{3} - \frac{\sqrt{2} \pi}{3}

    although my answer is different than yours but the same as the triple integral given by Galactus.
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