Suggested languages for you:

Americas

Europe

Q38E

Expert-verified
Found in: Page 380

### Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280

# To calculate the volume of the described solid (circular disk).

The volume of the circular disk is $$\frac{{16}}{3}{r^3}$$.

See the step by step solution

## Given data

The circular disk with radius $$r$$ and base $$S$$.

Parallel cross-sections perpendicular to the base are squares.

## Concept used of Volume

Volume is a scalar quantity expressed by the amount of three-dimensional space enclosed by a closed surface.

## Solve to find the volume

Consider that the Equation of circle $${x^2} + {y^2} = {r^2}$$ ….(1)

Rearrange Equation (1).

\begin{aligned}{l}{x^2} = {r^2} - {y^2}\\y = \sqrt {{r^2} - {x^2}} \end{aligned}

The cross-section of the circular disk as shown in Figure 1.

The region lies between $$a = - r$$ and $$b = r$$.

The expression to find the volume of the circular disk as shown below.

$$V = \int_a^b A (x)dx$$ …..(2)

Substitute $$- r$$ for a, r for $$b$$, and $$4\left( {{r^2} - {x^2}} \right)$$ for $$A(x)$$ in Equation (2).

\begin{aligned}{}V = \int_{ - r}^r 4 \left( {{r^2} - {x^2}} \right)dx\\ = 8\int_0^r {\left( {{r^2} - {x^2}} \right)} dx\\ = 8\left( {{r^2}x - \frac{{{r^3}}}{3}} \right)_0^r\\ = 8\left( {{r^2} \times r - \frac{{{r^3}}}{3} - 0} \right)\end{aligned}

Simplify further,

\begin{aligned}{}V = \frac{8}{3}\left( {3{r^3} - {r^3}} \right)\\ = \frac{{16}}{3}{r^3}\end{aligned}

Therefore, the volume of the circular disk is $$\frac{{16}}{3}{r^3}$$.