Question

1. Use double integration to find the volume of the solid bounded from above by the paraboloid z = 9x2 + y2, below by the plane z = 0, laterally by y=x^2 and y=x.

Answer 1

ANSWER The volume of the solid is $\frac{17}{35} \cong 0.4857$

EXPLANATION

Volume $=\iint_{D}\left (9x^2+y^2 \right )dydx$ ,

where

$D=\left \{ (x,y):0\leq x\leq 1, \, x^{2 }\leq y\leq x \right \}$

\iint_{D}\left (9x^2+y^2 \right )dydx=\int_{0}^{1} \left ( \int_{x^{2} }^{x}\left ( 9x^2+y^2 \right ) dy\right )dx\. Since $\int_{x^{2} }^{x}\left ( 9x^2+y^2 \right ) dy =\left [ 9x^{2}y+\frac{y^3}{3} \right ]_{y=x^2}^{y=x}=\left ( 9x^3-9x^4+\frac{x^3}{3}-\frac{x^{6} }{3} \right )$ , then $\int_{0}^{1}\left ( 9x^3-9x^4+\frac{x^3}{3}-\frac{x^{6} }{3} \right )dx=\left [ \frac{9x^{4}}{4}-\frac{9x^{5}}{5} +\frac{x^{4}}{12}-\frac{x^{7}}{21}\right ]_{0}^{1}= \frac{245-189-5}{ 105} =\frac{17}{35} \cong 0.4857$