How do you find the volume of the region enclosed by the curves $y = x$ $y=x$ , $y = - x$ $y=-x$ , and $x = 1$ $x=1$ rotated about $y = 1$ $y=1$ ?

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Answer 1 · 2015-08-27T18:05:23

The volume is $2 π$ $2pi$

Using the method of washers

Let the outer radius be $1 - (- x) = 1 + x$ $1-(-x)=1+x$

Let the inner radius be $1 - x$ $1-x$

The integral for the volume is

$π \int_{0}^{1} {(1 + x)}^{2} - {(1 - x)}^{2} d x$ $piint_0^1(1+x)^2-(1-x)^2dx$

$π \int_{0}^{1} (1 + 2 x + x^{2}) - (1 - 2 x + x^{2}) d x$ $piint_0^1(1+2x+x^2)-(1-2x+x^2)dx$

$π \int_{0}^{1} (1 + 2 x + x^{2} - 1 + 2 x - x^{2}) d x$ $piint_0^1(1+2x+x^2-1+2x-x^2)dx$

$π \int_{0}^{1} (4 x) d x$ $piint_0^1(4x)dx$

Integrating we get

$2 π x^{2}$ $2pix^2$

Evaluating from $0$ $0$ to $1$ $1$

$2 π {(1)}^{2} - 0 = 2 π$ $2pi(1)^2-0=2pi$

How do you find the volume of the region enclosed by the curves y=xy=x, y=−xy=-x, and x=1x=1 rotated about y=1y=1?