Question

Evaluate \int \:\int _D\left(x+y\right)^{-\frac{1}{2}}dA\: over the region 𝑥 − 2𝑦 ≤ 1 and 𝑥 ≥ 𝑦2 + 1

Answer 1

Evaluate $\iint _D\left(x+y\right)^{-\frac{1}{2}}dA$ : over the region 𝑥 − 2𝑦 ≤ 1 and 𝑥 ≥ 𝑦2 + 1

Solution:

$\begin{cases} x-2y=1 \\ x=y^2+1 \end{cases}$ $\begin{cases} x=2y+1 \\ x=y^2+1 \end{cases}$ $\begin{cases} x=1 & y=0 \\ x=5 & y=2 \end{cases}$

Region:

$\iint _D\left(x+y\right)^{-\frac{1}{2}}dA=\int_0^2dy\int_{y^2+1}^{2y+1}\frac{1}{\sqrt{x+y}}dx=$

$2\int_0^2(\sqrt{3y+1}-\sqrt{y^2+y+1})dy=$

$2\int_0^2(\sqrt{3y+1}-\sqrt{(y+\frac12)^2+\frac34})dy=$

$\left(\frac49(3y+1)^{\frac32}-(y+\frac12)\sqrt{(y+\frac12)^2+\frac34}-\frac 34\ln{(y+\frac12+\sqrt{(y+\frac12)^2+\frac34})}\right)|_0^2\approx$

0.747874.

Answer: $\iint _D\left(x+y\right)^{-\frac{1}{2}}dA\approx0.747874$ .