Q. 22 - 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. 

${\int }_0^4{\int }_0^{\sqrt{4y-y^2}}\frac{1}{\sqrt{x^2+y^2}}dxdy$

Step-by-Step Solution

Answer

${\int }_0^4{\int }_0^{\sqrt{4y-y^2}}\frac{1}{\sqrt{x^2+y^2}}dxdy=\mathrm{\pi }$

Step 1: 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, 

${\int }_0^4{\int }_0^{\sqrt{4y-y^2}}\frac{1}{\sqrt{x^2+y^2}}dxdy\Rightarrow {\int }_0^{\mathrm{\pi }/2}{\int }_2^{4}drd\theta$

Step 3: Calculate integral

$$\begin{gathered} I={\int }_0^{\mathrm{\pi }/2}{\int }_2^{4}drd\theta \\ I={\left[r\right]}_2^4{\int }_0^{\mathrm{\pi }/2}d\theta \\ I=2{\left[\theta \right]}_0^{\mathrm{\pi }/2} \\ I=2\times \frac{\mathrm{\pi }}{2} \\ I=\mathrm{\pi } \end{gathered}$$