Q. 23 - Calculus

Question

Using polar coordinates to evaluate iterated integrals: Evaluate the given iterated integrals by converting them to polar coordinates. Include a sketch of the region. 

${\int }_{-3}^3{\int }_{-\sqrt{9-x^2}}^{\sqrt{9-x^2}} \frac{x+2y}{x^2+y^2} dydx$

Step-by-Step Solution

Answer

${\int }_{-3}^3{\int }_{-\sqrt{9-x^2}}^{\sqrt{9-x^2}} \frac{x+2y}{x^2+y^2} dydx=0$

Step !: Draw the region

From the limits of integration, the region is shown below,  

Step 2: Convert into polar form

By using the below substitution,  

$$\begin{gathered} x=r\cos \theta \\ y=r\sin \theta \\ {x}^2+y^2=r^2 \\ dxdy=rdrd\theta \end{gathered}$$

The equivalent polar integral of the given integral is,  

${\int }_{-3}^3{\int }_{-\sqrt{9-x^2}}^{\sqrt{9-x^2}} \frac{x+2y}{x^2+y^2} dydx\Rightarrow {\int }_0^{2\mathrm{\pi }}{\int }_0^3 (\cos \theta +2\sin \theta ) drd\theta$

Step 3: Calculate the integral

$$\begin{gathered} I={\int }_0^{2\mathrm{\pi }}{\int }_0^3 (\cos \theta +2\sin \theta ) drd\theta \\ I={\int }_0^{2\mathrm{\pi }} (\cos \theta +2\sin \theta ) d\theta {\int }_0^{3}rdr \\ I={\left[\sin \theta -2\cos \theta \right]}_0^{2\mathrm{\pi }}{\int }_0^{3}rdr \\ I=\left[-2-(-2)\right]{\int }_0^{3}rdr \\ I=(-2+2){\int }_0^{3}rdr \\ I=0 \end{gathered}$$