Question

Sketch the region of integration for each of integrals in Exercises 57–60, and then evaluate the integral by converting to polar coordinates.

${\int }_0^1{\int }_{(\sqrt{3}/3)y}^{\sqrt{4-y^2}}\sqrt{x^2+y^2}dxdy$

Short answer

The required value of integral is ${\int }_0^1{\int }_{(\sqrt{3}/3)y}^{\sqrt{4-y^2}}\sqrt{x^2+y^2}dxdy=\frac{4\pi }{9}$

Step-by-step solution

Given information

The expression is${\int }_0^1{\int }_{(\sqrt{3}/3)y}^{\sqrt{4-y^2}}\sqrt{x^2+y^2}dxdy$

Calculation

The goal of this challenge is to draw the integration zone and then calculate the integral using polar coordinates.

The foundation is

$I={\int }_0^1{\int }_{(\sqrt{3}/3)y}^{\sqrt{4-y^2}}\sqrt{x^2+y^2}dxdy$

Here, $x=(\sqrt{3}/3)y$and $x=\sqrt{4-y^2},y=0$and $y=1$

The region of integration R is shown in the figure.

Substitute and at the lower limit of $x$

$$\begin{gathered} r\cos \theta =(\sqrt{3}/3)r\sin \theta \\ r\left(\cos \theta -\frac{1}{\sqrt{3}}\sin \theta \right)=0 \end{gathered}$$

This show $r=0$and $\tan \theta =\frac{1}{\sqrt{3}}$

$$\begin{gathered} \tan \theta =\frac{1}{\sqrt{3}} \\ \Rightarrow \theta =\frac{\pi }{6} \end{gathered}$$

$x=r\cos \theta$and $y=r\sin \theta$in the upper limit of $x$.

Substitute $y=r\sin \theta$in the lower limit of $y$

$$\begin{gathered} r\sin \theta =0 \\ \Rightarrow r=0,\theta =0 \end{gathered}$$

Thus, the limits of $r$are $r=0$and $r=2$and that of $\theta$are 0 and $\frac{\pi }{6}$.

$dxdy=rdrd\theta$

Therefore,

Integrate with respect to $x$first

Thus, the value of the integral is${\int }_0^1{\int }_{(\sqrt{3}/3)y}^{\sqrt{4-y^2}}\sqrt{x^2+y^2}dxdy=\frac{4\pi }{9}$