Question

Above the rectangle with vertices (0,0), (0,1),(2,0),(2,1), and below the surface ${\mathit{z}}^{\mathbf{2}}\mathbf{=}\mathbf{36}{\mathit{x}}^{\mathbf{2}}\mathbf{(}\mathbf{4}\mathbf{-}{\mathit{x}}^{\mathbf{2}}\mathbf{)}$

Short answer

The required solution is 16.

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$\mathbf{\int }{\mathbf{\int }}_{\mathbf{A}}\mathbf{f}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{)}\mathbf{dxdy}$

Calculation of the volume under the curve

The volume above the rectangle with vertices at (0,0),(0,1),(2,0), and (2,1), and under the plane z = 8 - x + y .below the surface$z^2=36x^2(4-x^2).$

First, integrate over z:

$$\begin{gathered} |={\int }_{x=0}^2{\int }_{y=0}^1\left[{\int }_{z=0}^{6x\sqrt{4-x^2}}dz\right]dxdy \\ ={\int }_0^{1}dy{\int }_0^{2}dx6x\sqrt{4-x^2} \end{gathered}$$

Further calculation of the volume of the bounded region

Now, integrate over x and y, such that,

$$\begin{gathered} |={\int }_0^{1}dy{\int }_0^{2}dx6x\sqrt{4-x^2} \\ =6{\int }_0^2\sqrt{4-x^2}d(4-x^2)\left(-\frac{1}{2}\right) \\ ={\overline{)-3\frac{2}{3}{(4-x^2)}^{3/2}}}_0^2 \\ =-2(0-4^{3/2)} \\ =2\times 2^3 \\ =16 \end{gathered}$$

Therefore, the value is 16