# A solid consists of a cone on top of a cylinder with a radius equal to that of the cone. The height of the cone is #36 # and the height of the cylinder is #8 #. If the volume of the solid is #48 pi#, what is the area of the base of the cylinder?

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Let (r) be the radius of the cylinder and the base of the cone.

The volume of the solid is the sum of the volumes of the cone and the cylinder:

[ V = V_{\text{cone}} + V_{\text{cylinder}} ]

[ V = \frac{1}{3}\pi r^2 h_{\text{cone}} + \pi r^2 h_{\text{cylinder}} ]

Given that ( h_{\text{cone}} = 36 ) and ( h_{\text{cylinder}} = 8 ), and ( V = 48\pi ), we have:

[ 48\pi = \frac{1}{3}\pi r^2 \cdot 36 + \pi r^2 \cdot 8 ]

Solving this equation for ( r ), we find:

[ r^2 = \frac{48\pi - 8\pi}{\frac{36}{3} + 8} ]

[ r^2 = \frac{40\pi}{12 + 8} ]

[ r^2 = \frac{40\pi}{20} ]

[ r^2 = 2\pi ]

[ r = \sqrt{2\pi} ]

The area of the base of the cylinder is ( A_{\text{cylinder}} = \pi r^2 ).

Substituting ( r = \sqrt{2\pi} ), we get:

[ A_{\text{cylinder}} = \pi (\sqrt{2\pi})^2 ]

[ A_{\text{cylinder}} = \pi (2\pi) ]

[ A_{\text{cylinder}} = 2\pi^2 ]

So, the area of the base of the cylinder is ( 2\pi^2 ).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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