Question

The value of the given integral by using polar coordinates.

Short answer

The value of the given iterated integral is \(\frac{{{a^5}}}{{15}}.\)

Step-by-step solution

Formula used

If\(f\)is a polar rectangle\(R\)given by\(0 \le a \le r \le b,\alpha \le \theta \le \beta \), where\(0 \le \beta - \alpha \le 2\pi \), then,

If\(g(x)\)is the function of\(x\)and\(h(y)\)is the function of\(y\)then,

\(\int_a^b {\int_c^d g } (x)h(y)dydx = \int_a^b g (x)dx\int_c^d h (y)dy.........(2)\)

Substitute \(x = r\cos \theta \) and \(y = r\sin \theta \) to convert the function into polar coordinates

\(\begin{aligned}{l}z = {x^2}y\\ = {(r\cos \theta )^2}r\sin \theta \\ = {r^3}{\cos ^2}\theta \sin \theta \end{aligned}\)

Moreover, from the given condition of \(x\) and \(y\), the value of \(r\) varies from 0 to \(a\) and the value of \(\theta \) varies from \(\frac{\pi }{2}\) to \(\pi \).

Therefore, by the equation (1), the value of the iterated integral becomes,

\begin{aligned}

\iint_{D} z d A &=\int_{\frac{\pi}{2}}^{\pi} \int_{0}^{a}\left(r^{3} \cos ^{2} \theta\right)(\sin \theta) r d r d \theta \\

&=\int_{\frac{\pi}{2}}^{\pi} \int_{0}^{a}\left(r^{4} \cos ^{2} \theta \sin \theta\right) d r d \theta \\

&=\left\{\int_{\frac{\pi}{2}}^{\pi} \cos ^{2} \theta \sin \theta d \theta\right\}\left\{\int_{0}^{a} r^{4} d r\right\}

\end{aligned}

Obtain the limit of \(r\) and \(\theta \)

\(\begin{aligned}{l}y = \sqrt {9 - {x^2}} \\{y^2} = 9 - {x^2}\\{x^2} + {y^2} = 9\\{r^2} = 9\\r = 3\end{aligned}\)

Therefore, the value of \(r\) varies from 0 to 3.

Since the value of \(y\) is positive, \(\theta \) lies in the first and second quadrants. Therefore, the value of \(\theta \) varies from 0 to \(\pi \).

Therefore, by equation (1), the value of the iterated integral is obtained as follows.

Let \(u = {r^2}\). Then \(du = 2rdr\).Also the limit of \(r\) varies from 0 to 9.

Therefore, the iterated integral is obtained as follows.

Integrate the function with respect to \(u\) and \(\theta \) by using equation (2)

Apply the limit of \(u\) and \(\theta \) as shown below.

Therefore, the value of the iterated integral is \(\frac{1}{2}\pi (1 - \cos 9)\).