Question

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

Short answer

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

Step-by-step solution

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

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.

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.

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

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 \(\frac{{16\pi }}{3}\).