Evaluating iterated integrals: Sketch the region determined by the limits of the given iterated integrals, and then evaluate the integrals. - Calculus

Question

Q. 7 ${\int }_0^1{\int }_{-\sqrt{1-y^2}}^{\sqrt{1+y^2}} \frac{x}{y+1} dxdy$

Step-by-Step Solution

Answer

${\int }_0^1{\int }_{-\sqrt{1-y^2}}^{\sqrt{1+y^2}} \frac{x}{y+1} dxdy=0$

Step 1: Draw the region

The region determined by the limits of the given integral is shown below,  

Step 2: Evaluate integral

Here a concept is used,

${\int }_{-a}^{a}f\left(x\right) dx=0, \mathrm{If} f\left(x\right) \mathrm{is} \mathrm{an} \mathrm{odd} \mathrm{function} \mathrm{i}.\mathrm{e}. f(-x)=-f\left(x\right)$

$$\begin{gathered} {\int }_0^1{\int }_{-\sqrt{1-y^2}}^{\sqrt{1+y^2}} \frac{x}{y+1} dxdy={\int }_0^1\frac{1}{y+1}\left[{\int }_{-\sqrt{1-y^2}}^{\sqrt{1+y^2}} x dx\right]dy \\ ={\int }_0^1\frac{1}{y+1}\left[0\right] dy \left[f\left(x\right)=x, \mathrm{is} \mathrm{an} \mathrm{odd} \mathrm{function}\right] \\ =0 \end{gathered}$$