Question

$\mathbf{\iint }{\mathbf{(}\mathbf{9}\mathbf{+}\mathbf{2}{\mathbf{y}}^{\mathbf{2}}\mathbf{)}}^{\mathbf{-}\mathbf{1}}$ over the quadrilateral with vertice$\left(1,3\right)\mathbf{,}\left(3,3\right)\mathbf{,}\left(2,6\right)\mathbf{,}\left(6,6\right)$

Short answer

The required solution is$\frac{\mathrm{In}3}{6}$

Step-by-step solution

Definition of double integral

The double integral of$\mathit{f}\left(x,y\right)$ over the areain the$\left(x,y\right)$plane as the limit of this sum, and we write it as${\mathbf{\iint }}_{\mathbf{A}}\mathit{f}\left(x,y\right)\mathit{d}\mathit{x}\mathit{d}\mathit{y}$.

 Step 2: Drawing the area bounded by the curve

The area is bounded by the quadrilateral with vertices$\left(1,3\right),\left(3,3\right),\left(2,6\right),\left(6,6\right)$.

Calculation of the area under the curve

First, integrate over x:

$$\begin{gathered} {\iint }_A\frac{dydx}{9+2y^2}={\int }_{y=3}^6\left[{\int }_{x=y/3}^{y}dx\right]\frac{dy}{9+2y^2} \\ ={\int }_3^6\frac{2}{3}y\frac{dy}{2y^2} \end{gathered}$$

Further calculation of the area of the bounded region

Now, integrate over y:

$$\begin{gathered} \frac{2}{3}{\int }_3^6\frac{ydy}{9+2y^2}=\frac{1}{6}{\int }_3^6\frac{\left(d9+2y^2\right)}{9+2y^2} \\ =\frac{1}{6}In{\overline{)\left(9+2y^2\right)}}_3^6 \\ =\frac{In3}{6} \end{gathered}$$

Therefore, the value is$\frac{In3}{6}$