How do you use the shell method to set up and evaluate the integral that gives the volume of the solid generated by revolving the plane region $y = 1$ $y=1$ , $y = x^{2}$ $y=x^2$ , and $x = 0$ $x=0$ rotated about the line $y = 2$ $y=2$ ?

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Answer 1 · 2015-09-22T21:13:02

$\frac{28 π}{15}$ $(28pi)/15$ cubic units

Since we are revolving around a horizontal line using the method of shells we will integrate with respect to $y$ $y$ .

We are bounded by the $y$ $y$ axis, the horizontal line $y = 1$ $y=1$ , and the function $y = x^{2}$ $y=x^2$

Solve $y = x^{2}$ $y=x^2$ for $x$ $x$

$x = \sqrt{y}$ $x=sqrt(y)$

We are in quadrant $I$ $I$ so we do not have to worry about the negative square root.

Our representative cylinder height is our function $\sqrt{y}$ $sqrt(y)$

Our representative radius is $2 - y$ $2-y$ over the interval $0 \leq y \leq 1$ $0<=y<=1$

The integral for the volume is

$2 π \int_{0}^{1} (2 - y) (y^{\frac{1}{2}}) d y$ $2piint_0^1(2-y)(y^(1/2))dy$

$2 π \int_{0}^{1} 2 y^{\frac{1}{2}} - y^{\frac{3}{2}} d y$ $2piint_0^1 2y^(1/2)-y^(3/2)dy$

Integrating we get

$2 π [\frac{4}{3} y^{\frac{3}{2}} - \frac{2}{5} y^{\frac{5}{2}}]$ $2pi[4/3y^(3/2)-2/5y^(5/2)]$

Evaluating we get

$2 π [\frac{4}{3} - \frac{2}{5} - 0]$ $2pi[4/3-2/5-0]$

$2 π [\frac{20}{15} - \frac{6}{15}] = 2 π [\frac{14}{15}] = \frac{28 π}{15}$ $2pi[20/15-6/15]=2pi[14/15]=(28pi)/15$ cubic units

How do you use the shell method to set up and evaluate the integral that gives the volume of the solid generated by revolving the plane region y=1y=1, y=x2y=x^2, and x=0x=0 rotated about the line y=2y=2?