Q. 21 - 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.

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

Step-by-Step Solution

Answer

$\frac{\mathrm{\pi }({\mathrm{e}}^4-1)}{4}$

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,

${\int }_0^2{\int }_0^{\sqrt{4-y^2}}{e}^{x^2+y^2} dxdy\Rightarrow {\int }_0^{\mathrm{\pi }/2}{\int }_0^2{e}^{r^2} rdrd\theta$

Step 3: Calculate integral

$$\begin{gathered} I={\int }_0^{\mathrm{\pi }/2}\left[{\int }_0^2{e}^{r^2} rdr\right]d\theta \\ I={\int }_0^{\mathrm{\pi }/2}{\left[\frac{1}{2}{e}^{r^2}\right]}_0^{2}d\theta \\ I=\frac{1}{2}(e^4-1){\int }_0^{\mathrm{\pi }/2}d\theta \\ I=\frac{1}{2}(e^4-1){\left[\theta \right]}_0^{\mathrm{\pi }/2} \\ I=\frac{\mathrm{\pi }({\mathrm{e}}^4-1)}{4} \end{gathered}$$