Question

(a) The base of a solid is a square with vertices located at (1,0),(0,1),(-1,0)and (0.-1)Each cross- section perpendicular to the x-axis is a semicircle. Find the volume of the solid.

(b) Show that by cutting the solid of part (a), we can rearrange it to from a cone. Thus compute its volume more simply.

Short answer

(a) The volume of the solid is \(\frac{\pi }{3}\)

(b) The volume of the solid of part (a) can be computed more simply by first cutting the solid and rearranging it to from a cone.

Step-by-step solution

Solution of the volume of the solid

By using the formula for volume

Let applied the values in formula

\(V = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n A \left( {x_i^*} \right)\Delta x = \int_a^b A (x)dx\)

\(\begin{aligned}A &= \frac{1}{2}\pi {r^2}\\ &= \frac{1}{2}\pi {(x + 1)^2}\\V &= 2\int_{ - 1}^0 {\frac{1}{2}} \pi {(x + 1)^2}dx\\V &= \pi \int_{ - 1}^0 {\left( {1 + 2x + {x^2}} \right)} dx\\V &= \pi \left( {x + {x^2} + \frac{{{x^3}}}{3}} \right)_{ - 1}^0\\V &= \pi \left( { - 1 + 1 + \frac{1}{3}} \right)\\ &= \frac{\pi }{3}\end{aligned}\)

Solution of the solid can be computed by more simply

By using the formula for volume of a cone

\(\begin{aligned}V &= \frac{1}{3}\pi {r^2}h\\r &= 1\\h &= 1\\V &= \frac{1}{3}\pi ({1^2})(1)\\V &= \frac{\pi }{3}\end{aligned}\)

The volume of the solid of part (a) can be computed more simply by first cutting the solid and rearranging it to from a cone.