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Thread: Triple Integral

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    Senior Member polymerase's Avatar
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    Triple Integral

    Evaluate $\displaystyle \int\int_E\int \sqrt{x^2+y^2+z^2}\;dV $where E is the solid bounded below by the conic surface $\displaystyle 3z^2=x^2+y^2,z\ge 0$ and bounded above by the spherical surface $\displaystyle x^2+y^2+z^2=5$.

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  2. #2
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    Quote Originally Posted by polymerase View Post
    Evaluate $\displaystyle \int\int_E\int \sqrt{x^2+y^2+z^2}\;dV $where E is the solid bounded below by the conic surface $\displaystyle 3z^2=x^2+y^2,z\ge 0$ and bounded above by the spherical surface $\displaystyle x^2+y^2+z^2=5$.

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    The upper surface is $\displaystyle z = \sqrt{5-x^2-y^2}$ and lower surface is $\displaystyle z = \frac{1}{\sqrt{3}}\sqrt{x^2+y^2}$. You need to know the region over which you are integrating. Note if $\displaystyle x^2+y^2+z^2 = 5 \implies 3x^2+3y^2+3z^2 = 15$ so $\displaystyle 4x^2+4y^2 = 15\implies x^2+y^2 = \frac{15}{4}$.

    Thus, we get, $\displaystyle \iint_A \int_{(1/3)\sqrt{x^2+y^2}}^{\sqrt{5-x^2-y^2}} f(x,y,z) dz dA$ where $\displaystyle A$ is the disk $\displaystyle x^2+y^2 \leq \frac{15}{4}$.

    Now use Polar change of variable.
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