How do you find the volume of the region enclosed by the curves #y=x#, #y=-x#, and #x=1# rotated about #y=1#?
The volume is
Using the method of washers
The integral for the volume is
Integrating we get
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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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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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