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 \(4\pi \).

Step-by-step solution

Given data

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

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 \( - \frac{\pi }{2}\) and \(\frac{\pi }{2}\), the region is the semicircle in the \(xy\)-plane with radius \(2\) . \(z\)varies from 0 to \({r^2}\). 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.\(\begin{array}{c}\int_{\frac{x}{2}}^{\frac{\pi }{2}} {\int_0^2 {\int_0^{{r^2}} r } } dzdrd\theta &=& \int_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {\int_0^2 ( rz)_0^{{r^2}}} drd\theta \\ &=& \int_{\frac{x}{2}}^{\frac{\pi }{2}} {\int_0^2 {\left( {r\left( {{r^2}} \right) - r(0)} \right)} } drd\theta \\ &=& \int_{\frac{\pi }{2}}^{\frac{\pi }{2}} {\int_0^2 {\left( {{r^3} - 0} \right)} } drd\theta \\ &=& \int_{\frac{\pi }{2}}^{\frac{\pi }{2}} {\int_0^2 {{r^3}} } drd\theta \end{array}\)

Integrate the integral with respect to \(r\)

Integrate the given iterated integral with respect to \(r\) and apply the limit of it.

\(\begin{array}{c}\int_{\frac{x}{2}}^{\frac{\pi }{2}} {\int_0^2 {\int_0^{{r^2}} r } } dzdrd\theta &=& \int_{\frac{x}{2}}^{\frac{\pi }{2}} {\left( {\frac{{{r^4}}}{4}} \right)_0^2} d\theta \\ &=& \int_{\frac{{ - \pi }}{2}}^{\frac{\pi }{2}} {\left( {\frac{{{2^4}}}{4} - \frac{{{0^4}}}{4}} \right)} d\theta \\ &=& \int_{\frac{{ - x}}{2}}^{\frac{\pi }{2}} {\left( {\frac{{16}}{4} - 0} \right)} d\theta \\ &=& \int_{\frac{{ - x}}{2}}^{\frac{\pi }{2}} 4 d\theta \end{array}\)

Integrate the integral with respect to \(\theta \)

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

\(\begin{array}{c}\int_{\frac{{ - \pi }}{2}}^{\frac{\pi }{2}} {\int_0^2 {\int_0^{{r^2}} r } } dzdrd\theta &=& (4\theta )_{ - \frac{x}{2}}^{\frac{x}{2}}\\ &=& 4\left( {\frac{\pi }{2} - \left( { - \frac{\pi }{2}} \right)} \right)\\ &=& 4\left( {\frac{\pi }{2} + \frac{\pi }{2}} \right)\\ &=& 4\pi \end{array}\)

Thus, the value of the given iterated integral is \(4\pi \).