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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

William Abdalla

#pi# cubic units.

The volume of this torus-like solid is

#pi int((1+x)^2-(1-x)^2) dx#, from x = 9 to x =1

#=pi int 2x dx#, for the limits

#=pi[x^2]#, between x = 0 and x = 1

#pi# cubic units.

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Answer 2

Samantha Adkison

To find the volume of the region enclosed by the curves ( y = x ), ( y = -x ), and ( x = 1 ) rotated about ( y = -1 ), you can use the method of cylindrical shells. The integral setup for this problem is as follows:

[ V = 2\pi \int_{0}^{1} (x+1)(1-x) , dx ]

Solving this integral will give you the volume of the region.

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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.