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 $y = x$ $y=x$ , $0 \leq x \leq 1$ $0<=x<=1$ rotated about the x-axis?

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Answer 1 · 2016-05-30T15:05:04

Volume $V = \frac{π}{3}$ $V=pi/3" "$ cubic units

Using the cylindrical shell method. The differential is

$d V = 2 π \cdot r \cdot h \cdot d r$ $dV=2pi*r*h*dr$

$d V = 2 \cdot π \cdot y \cdot (1 - x) \cdot d y$ $dV=2*pi*y*(1-x)*dy$

but $x = y$ $x=y$ , therefore

$d V = 2 \cdot π \cdot y \cdot (1 - y) \cdot d y$ $dV=2*pi*y*(1-y)*dy$

our limits for x are $0 \to 1$ $0 rarr 1$
our limits for y are $0 \to 1$ $0 rarr 1$

We solve the volume by integrating both sides with limits $y = 0$ $y=0$ to $y = 1$ $y=1$

$\int d V = 2 \cdot π \cdot \int y \cdot (1 - y) \cdot d y$ $int dV=2*pi*int y*(1-y)*dy$

$V = 2 \cdot π \cdot \int_{0}^{1} (y - y^{2}) \cdot d y$ $V=2*pi*int_0^1 (y-y^2)*dy$

$V = 2 \cdot π \cdot {[\frac{y^{2}}{2} - \frac{y^{3}}{3}]}_{0}^{1}$ $V=2*pi* [y^2/2-y^3/3]_0^1$

$V = 2 \cdot π \cdot [\frac{1^{2}}{2} - \frac{1^{3}}{3} - (\frac{0^{2}}{2} - \frac{0^{3}}{3})]$ $V=2*pi* [1^2/2-1^3/3-(0^2/2-0^3/3)]$

$V = 2 \cdot π \cdot [\frac{1}{2} - \frac{1}{3} - 0]$ $V=2*pi* [1/2-1/3-0]$

$V = 2 \cdot π \cdot \frac{1}{6}$ $V=2*pi* 1/6$

$V = \frac{π}{3}$ $V=pi/3" "$ cubic units

God bless....I hope the explanation is useful.

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=xy=x, 0≤x≤10<=x<=1 rotated about the x-axis?