Question

Evaluate the iterated integral by using polar coordinates.

Short answer

The value of the iterated integral is \(\frac{{2\sqrt 2 }}{3}.\)

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

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

Obtain the limit of \(r\) and \(\theta \) as follows.

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

Therefore, the value of \(r\) varies from 0 to \(\sqrt 2 \).

The value of \(\theta \) varies from 0 to \(\frac{\pi }{4}\).

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

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

\begin{aligned} \iint_{D} z d A &=\int_{0}^{\frac{\pi}{4}}(\cos \theta+\sin \theta) d \theta \int_{0}^{\sqrt{2}} r^{2} d r \\ &=(\sin \theta-\cos \theta)_{0}^{\frac{\pi}{4}}\left(\frac{r^{3}}{3}\right)_{0}^{\sqrt{2}} \end{aligned}