Question

Under the surface z = 1 /(y+2) , and over the area bounded by and y=x ${\mathbf{y}}^{\mathbf{2}}\mathbf{+}\mathbf{x}\mathbf{=}\mathbf{2}$.

Short answer

The required solution is$\frac{9}{2}$

Step-by-step solution

Definition of double integral

The double integral of f(x,y)over the areaA in the (x,y)plane as the limit of this sum, and we write it as$\mathbf{\int }{\mathbf{\int }}_{\mathbf{A}}\mathbf{f}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{)}\mathbf{dxdy}$

Drawing the area bounded by the curve inthe plane

The area is bounded by the graph y=x and $y^2+x=2\mathrm{in}(x,y)$the plane

Intersecting points identification

From the graph the intersections of the two curves are:

$$\begin{gathered} y=-y^2 \\ {y}_{12}=\frac{-1\pm \sqrt{1+8}}{2} \\ {y}_{12}=(-2,1) \end{gathered}$$

Calculation of the volume under the curve

First, integrate over z and x, such that,

$$\begin{gathered} |={\int }_{y=2}^{1}dy{\int }_{x=y}^{2-y^2}\left[{\int }_{z=0}^{1/(y+2)}dz\right]dx \\ ={\int }_{-2}^{1}dy{\int }_x^{2-y^2}\frac{dx}{y+2} \\ ={\int }_{-2}^1{\overline{)\frac{dx}{y+2}x}}_y^{2-y^2} \\ ={\int }_{-2}^1\frac{dx}{y+2}(2-y^2-y) \end{gathered}$$

Further calculation of the volume of the bounded region

Now, integrate over y:

$$\begin{gathered} |={\int }_{-2}^1\frac{dy}{y+2}(y+2\left)\right(y-1\left)\right(-1) \\ ={\int }_{-2}^1(y-1)dy \\ ={\overline{)-\left[\frac{1}{2}{y}^2-y\right]}}_{-2}^1 \\ =\frac{9}{2} \end{gathered}$$

Therefore, the value is$\frac{9}{2}$