Question

Above the triangle with vertices (0,0),(2,0), and (2,1), and below the paraboloid $\mathrm{z}=24-{\mathrm{x}}^2-{\mathrm{y}}^2$.

Short answer

The required solution is$\frac{131}{6}$

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 triangle with vertices at (0,0),(2,0),(2,1) and below the paraboloid $z=24-x^2-y^2$.

First, integrate over z:

$$\begin{gathered} |={\int }_{\mathrm{x}=0}^2{\int }_{\mathrm{y}=0}^{\mathrm{x}/2}\left[{\int }_{\mathrm{z}=0}^{24-{\mathrm{x}}^2-{\mathrm{y}}^2}\mathrm{dz}\right]\mathrm{dydx} \\ ={\int }_0^2\mathrm{dx}{\int }_0^{\mathrm{x}/2}(24-{\mathrm{x}}^2-{\mathrm{y}}^2) \end{gathered}$$

Further calculation of the volume of the bounded region

Now, integrate over:

$$\begin{gathered} |={\int }_0^2\mathrm{dx}{\overline{)\left(24\mathrm{y}-{\mathrm{x}}^2\mathrm{y}-\frac{1}{3}{\mathrm{y}}^3\right)}}_0^{\mathrm{x}/2} \\ ={\int }_0^2\left(12\mathrm{x}-\frac{{\mathrm{x}}^3}{2}-\frac{{\mathrm{x}}^3}{24}\right)\mathrm{dx} \end{gathered}$$

Further calculation of the volume of the bounded region

Now, integrate over x:

$$\begin{gathered} |={\overline{)\left[6{\mathrm{x}}^2-\frac{{\mathrm{x}}^3}{8}-\frac{{\mathrm{x}}^3}{96}\right]}}_0^2 \\ =\frac{131}{6} \end{gathered}$$

Therefore, the value is$\frac{131}{6}$