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

Answer 1

The volume is #2pi#

Using the method of washers

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

The integral for the volume is

#piint_0^1(1+x)^2-(1-x)^2dx#
#piint_0^1(1+2x+x^2)-(1-2x+x^2)dx#
#piint_0^1(1+2x+x^2-1+2x-x^2)dx#
#piint_0^1(4x)dx#

Integrating we get

#2pix^2#
Evaluating from #0# to #1#
#2pi(1)^2-0=2pi#
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Answer 2

To find the volume of the region enclosed by the curves (y=x), (y=-x), and (x=1) rotated about (y=1), you use the method of cylindrical shells.

The integral setup is:

[ V = 2\pi \int_0^1 (1 - x)(x - (-x)) , dx ]

[ V = 2\pi \int_0^1 (1 - x)(2x) , dx ]

[ V = 4\pi \int_0^1 (x - x^2) , dx ]

[ V = 4\pi \left[\frac{x^2}{2} - \frac{x^3}{3}\right]_0^1 ]

[ V = 4\pi \left(\frac{1}{2} - \frac{1}{3}\right) ]

[ V = \frac{4\pi}{6} = \boxed{\frac{2\pi}{3}} ]

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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