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Q38E

Expert-verifiedFound in: Page 380

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 \(\frac{{16}}{3}{r^3}\).

The circular disk with radius \(r\) and base \(S\).

Parallel cross-sections perpendicular to the base are squares.

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

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 \(\frac{{16}}{3}{r^3}\).

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