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 $y = 1 + x^{2}$ $y = 1 + x^2$ , $y = 0$ $y = 0$ , $x = 0$ $x = 0$ , $x = 2$ $x = 2$ rotated about the line $x = 4$ $x=4$ ?

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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 $y = 1 + x^{2}$ $y = 1 + x^2$ , $y = 0$ $y = 0$ , $x = 0$ $x = 0$ , $x = 2$ $x = 2$ rotated about the line $x = 4$ $x=4$ ?

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Answer 1 · 2017-09-28T04:17:11

$\frac{76 π}{3}$ $(76pi)/3$

Explanation:

This is a graph of the region that will be revolved around the vertical line $x = 4$ $x =4$ (not pictured).

graph{(y-1-x^2)(y)( sqrt(2-x) )(sqrt(x)) / (sqrt(2-x))/(sqrt(x))<=0 [0, 6, -1.51, 6.39]}

Recall the general form of the volume of a solid of revolution using the shells method when you are revolving about a vertical line:

$V = \int_{a}^{b} 2 π \cdot r a d i u s \cdot f (x) d x$ $V = int_a^b 2pi * radius * f(x)" " dx$

The hardest conceptual part here is the radius. Since the axis of revolution is $x = 4$ $x = 4$ , the form of the radius is the expression $4 - x$ $4 - x$ .

(Why? Draw a segment from $x = 4$ $x = 4$ on the x-axis to $x = 1$ $x = 1$ on the x-axis. How long is that segment? It is $4 - 1 = 3$ $4 - 1 = 3$ .

Thus:

$V = \int_{0}^{2} 2 π \cdot (4 - x) \cdot (1 + x^{2}) d x$ $V = int_0^2 2pi * (4-x) * (1 + x^2) dx$
$= 2 π \int_{0}^{2} (4 + 4 x^{2} - x - x^{3}) d x$ $= 2pi int_0^2 (4 + 4x^2 - x - x^3) dx$
$= 2 π (4 x + \frac{4}{3} x^{3} - \frac{1}{2} x^{2} - \frac{1}{4} x^{4}) {∣ ∣ ∣}_{0}^{2}$ $=2pi (4x + 4/3x^3 - 1/2x^2 - 1/4x^4)|_0^2$
$= 2 π (4 \cdot 2 + \frac{4}{3} \cdot 2^{3} - \frac{1}{2} \cdot 2^{2} - \frac{1}{4} \cdot 2^{4})$ $=2pi(4*2 + 4/3 * 2^3 - 1/2 * 2^2 - 1/4 * 2^4)$
$= 2 π (8 + \frac{32}{3} - 2 - 4) = 2 π (2 + \frac{32}{3}) = 2 π (\frac{38}{3}) = \frac{76 π}{3}$ $=2pi(8 + 32/3 - 2 - 4) = 2pi(2 + 32/3) = 2pi(38/3)=(76pi)/3$

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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 $y = 1 + x^{2}$ $y = 1 + x^2$ , $y = 0$ $y = 0$ , $x = 0$ $x = 0$ , $x = 2$ $x = 2$ rotated about the line $x = 4$ $x=4$ ?

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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+x2y = 1 + x^2, y=0y = 0, x=0x = 0, x=2x = 2 rotated about the line x=4x=4?

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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 $y = 1 + x^{2}$ $y = 1 + x^2$ , $y = 0$ $y = 0$ , $x = 0$ $x = 0$ , $x = 2$ $x = 2$ rotated about the line $x = 4$ $x=4$ ?