Pr. #10) Using triple integrals, find the volume of the solid contained in the first octant, bounded above and below by the cone $z^2 = \frac{1}{3}(x^2 + y^2)$, and to the side by the sphere $x^2 + y^2 + z^2 = a > 0$.
Added by Steve D.
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Step 1
First, let's set up the limits of integration for the triple integral. Since the cone is contained in the first octant, we can set the limits of integration as follows: - For x: 0 to 2 - For y: 0 to 2-x - For z: 0 to √(x^2 + y^2) Show more…
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