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"Please use Calculus to solve:

Find the volume of a solid whose base is a circle of radius r and each cross-section perpendicular to the base is an isosceles triangle with height h."

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Final answer:

The volume of the solid is 2πr^2 * h, where r is the radius of the circular base and h is the height of the isosceles triangles.

Step-by-step explanation:

To find the volume of the solid, we need to consider each cross-section and calculate its area. Since each cross-section is an isosceles triangle, we can use the formula for the area of a triangle, which is A = (1/2)base x height.

In this case, the base is the diameter of the circular base, which is 2r, and the height of the triangle is h. So the area of each cross-section is A = (1/2)(2r)(h) = r * h.

To find the volume, we integrate the area function over the range of the base. Since the base is a circle of radius r, the range is from 0 to 2πr. So the volume is V = ∫(0 to 2πr) (r * h) dx = ∫(0 to 2πr) r * h dx = r * h * ∫(0 to 2πr) 1 dx = r * h * [x] from 0 to 2πr = r * h * (2πr - 0) = 2πr^2 * h.

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