Q. 43 - Calculus

Question

Find the volumes of the solids described in Exercises 41–44. The solid bounded above by the paraboloid with equation$z=8-x^2-y^2$and bounded below by the rectangle$R=\left\{\right(x,y\left)\right|1\le x\le 2$and $0\le y\le 2\}$ in the xy-plane.

Step-by-Step Solution

Answer

The volume of the described solid is:$\frac{26}{3} \mathrm{units}$

Step 1. Given Information

Solid bounded above by a function,$z=f(x,y)=8-x^2-y^2$

Solid bounded below by region,$R=\left\{\right(x,y\left)\right|1\le x\le 2$and $0\le y\le 2\}$

Step 2. Finding the volume of given solid

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

$$\begin{gathered} {\int }_0^2{\int }_1^2\left(8-{\mathrm{x}}^2-{\mathrm{y}}^2\right) \mathrm{dx} \mathrm{dy} \\ ={\int }_0^2{\overline{)\left(8\mathrm{x}-\frac{{\mathrm{x}}^3}{3}-{\mathrm{y}}^2\mathrm{x}\right)}}_1^2 \mathrm{dy} \\ ={\int }_0^2\left(8(2-1)-\frac{2^3-1^3}{3}-{\mathrm{y}}^2(2-1)\right) \mathrm{dy} \\ ={\int }_0^2\left(8-\frac{7}{3}-{\mathrm{y}}^2\right) \mathrm{dy} \\ ={\int }_0^2\left(\frac{17}{3}-{\mathrm{y}}^2\right) \mathrm{dy} \\ ={\overline{)\left(\frac{17\mathrm{y}}{3}-\frac{{\mathrm{y}}^3}{3}\right)}}_0^2 \\ =\left(\frac{17(2-0)}{3}-\frac{2^3-0^3}{3}\right) \\ =\frac{34}{3}-\frac{8}{3} \\ =\frac{26}{3} \end{gathered}$$