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Found in: Page 734

### Essential Calculus: Early Transcendentals

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

# To sketch the solid whose volume is given by the iterated integral and evaluate it.

The value of the given iterated integral is $$\frac{{16\pi }}{3}$$.

See the step by step solution

## Step 1: Given data

The region $$D$$ is $$\left\{ {(r,\theta ,z)\mid 0 \le r \le 2,0 \le \theta \le 2\pi ,0 \le z \le r} \right\}$$.

## Step 2: Formula used of Integration

Integration $$\mathop \smallint \nolimits^b f(x)dx = f(b) - f(a)$$

## Step 3: Draw the sketch

Since $$r$$ varies between $$0$$and $$2$$ and $$\theta$$ varies between $$0$$ and $$2\pi$$, the region is the circle in the $$xy$$-plane with radius $$2$$ . $$z$$varies from 0 to $$r$$. So, the outline sketch of the given region $$D$$ is given below in the Figure 1.

## Step 4: Integrate the integral with respect to $$z$$

Integrate the given iterated integral with respect to $$z$$ and apply the limit of it. $$\begin{array}{c}\int_0^2 {\int_0^{2\pi } {\int_0^r r } } dzd\theta dr &=& \int_0^2 {\int_0^{2\pi } {(rz)} _0^rd\theta } dr\\ &=& \int_0^2 {\int_0^{2\pi } {(r(r) - r(0))} _0^rd\theta } dr\\ &=& \int_0^2 {\int_0^{2\pi } {\left( {{r^2} - 0} \right)} } d\theta dr\\ &=& \int_0^{2\pi } {\int_0^{{r^2}} d } \theta dr\end{array}$$

## Step 5: Integrate the integral with respect to $$r$$

Integrate the given iterated integral with respect to $$r$$ and apply the limit of it.

$$\begin{array}{c}\int_0^2 {\int_0^{2\pi } {\int_0^r r } } dzd\theta dr &=& \int_0^2 {\left( {{r^2}\theta } \right)_0^{2\pi }} dr\\ &=& \int_0^2 {\left( {{r^2}(2\pi ) - {r^2}(0)} \right)} dr\\ &=& \int_0^2 {\left( {2\pi {r^2} - 0} \right)} dr\\ &=& \int_0^2 2 \pi {r^2}dr\end{array}$$

## Step 6: Integrate the integral with respect to $$\theta$$

Integrate the given iterated integral with respect to $$\theta$$ and apply the limit of it.

$$\begin{array}{c}\int_0^2 {\int_0^{2\pi } {\int_0^r r } } dzd\theta dr &=& \left( {2\pi \frac{{{r^3}}}{3}} \right)_0^2\\ &=& 2\pi \left( {\frac{{{2^3}}}{3} - \frac{{{0^3}}}{3}} \right)\\ &=& 2\pi \left( {\frac{8}{3} - 0} \right)\\ &=& \frac{{16\pi }}{3}\end{array}$$

Thus, the value of the given iterated integral is $$\frac{{16\pi }}{3}$$.