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Set up a triple integral for the volume of the solid. The solid that is the common interior below the sphere \(x^{2}+y^{2}+z^{2}=80\) and above the paraboloid \(z=\frac{1}{2}\left(x^{2}+y^{2}\right)\)

Short Answer

Expert verified
The triple integral that represents the volume of the solid within the sphere and above the paraboloid is \(\int_0^{2蟺}\int_0^{arccos\sqrt{1 - 蟻^{2}}}\int_0^{\sqrt{80}} 蟻^{2}sin(蠁)d蟻d蠁d胃\).

Step by step solution

01

Convert the given equations into spherical coordinates

In spherical coordinates, the sphere \(x^{2}+y^{2}+z^{2}=80\) becomes \(蟻^{2}=80\) or \(蟻= \sqrt{80}\). The paraboloid \(z=\frac{1}{2}\left(x^{2}+y^{2}\right)\) becomes \(\cos(蠁)=蟻sin(蠁)\) or \(蠁=\arccos\sqrt{1 - 蟻^{2}}\). The angle \(胃\) will range over all values, which is [0, 2蟺]. The volume element in spherical coordinates is \(dV = 蟻^{2}sin(蠁)d蟻d蠁d胃\).
02

Set up the integral

Now we set up the triple integral. The limits of the integral for \(蠁\) are determined by the sphere and the paraboloid, which are: \([0, \arccos\sqrt{1 - 蟻^{2}}]\). The limits for \(胃\) are [0, 2蟺] as \(胃\) covers all the angles in a circle around the positive z-axis. The limits for \(蟻\) are [0, \sqrt{80}] as \(蟻\) ranges from the origin to surface of the sphere. We integrate the volume element \(\int^{\sqrt{80}}_\sigma \int^{2蟺}_0 \int^{arccos\sqrt{1 - 蟻^{2}}} _0 蟻^{2}sin(蠁)d蠁d胃d蟻\).
03

Simplify the integral

Simplify the triple integral: in this problem we don't actually carry out the integration because the task only asks for setting up the integral for the volume of the solid.

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