Question

Find the volumes of the solids described in Exercises 41–44. The solid bounded above by the hyperboloid with equation$z=x^2-y^2$and bounded below by the square with vertices:$(2,2,-4),(2,-2,-4),(-2,-2,-4),(-2,2,-4).$

Short answer

The volume of the described solid is: 0

Step-by-step solution

Step 1. Given Information  

The solid bounded above by the hyperboloid, $z=x^2-y^2$

The solid bounded below by the square with vertices:

$(2,2,-4),(2,-2,-4),(-2,-2,-4),\mathrm{and}(-2,2,-4).$

The region is:$R=\left\{\right(x,y\left)\right|-2\le x\le 2$and$-2\le y\le 2\}$

Step 2. Finding the volume of given solid

The double integral to find the volume of the given solid is given by,

$$\begin{gathered} {\int }_{-2}^2{\int }_{-2}^2\left({\mathrm{x}}^2-{\mathrm{y}}^2\right)dydx \\ ={\int }_{-2}^2{\overline{)\left({\mathrm{x}}^2\mathrm{y}-\frac{{\mathrm{y}}^3}{3}\right)}}_{-2}^{2}dx \\ ={\int }_{-2}^2\left(x^2(2-(-2)-\frac{2^3-{(-2)}^3}{3}\right)dx \\ ={\int }_{-2}^2\left(4x^2-\frac{16}{3}\right)dx \\ ={\overline{)\left(\frac{4x^3}{3}-\frac{16\mathrm{x}}{3}\right)}}_{-2}^2 \\ =\frac{4(2^3-{(-2)}^3)}{3}-\frac{16(2-(-2\left)\right)}{3} \\ =\frac{64}{3}-\frac{64}{3} \\ =0 \end{gathered}$$