How do you find the volume of the solid generated by revolving the region bounded by the curves y = x² and y =1 rotated about the y=-2?

Answer 1

See the explanation below.

Here is a picture of the region and a vertical slice.

The picture is set up to use washers (disks).
Thickness is #dx#
#x# values go from #-1# to #1#
the radius of the larger washer is the greater #y# minus #-2# (the line we are revolving about is #y=-2#)
#R = 1-(-2) = 3#
the radius of the smaller washer is the lesser #y# minus #-2#
#r=x^2-(-2) = x^2+2#

The representative slice has volume #pi(R^2-r^2)dx#.

We need to evaluate the integral

#int_-1^1 pi((3^2-(x^2+2)^2)dx=pi int_-1^1(5-4x^2-x^4)dx#

# = 104/15pi#

Shells Method
If we had taken a slice horizontally:

This is set up to use cylindrical shells of thickness #dy#

The volume of each shell is #2pi("radius")("height")("thickness")# with thickness #dy#, that is #2pi*rh*dy#
In the region, the #y# values go from #0# to #1#.
The radius of the representative shell is #y+2# (dotted line in picture)
The height goes from the greater #x# (the one on the right) to the lesser #x# (the one on the left).
We need to rewrite the boundary as functions of #y# instead of #x#. #y=x^2# becomes the two functions #x=-sqrty# (on the left) and #x=sqrty# (on the right). The height of the shell is #sqrty-(-sqrty) = 2sqrty#

The representative shell has volume #2pi(y+2)(2sqrty)dy#.

The solid has volume

#V = int_0^1 2pi(y+2)(2sqrty)dy = 4piint_0^1 (y^(3/2)+2y^(1/2))dy#

# =4pi(26/15) = 104/15 pi#

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Answer 2

To find the volume of the solid generated by revolving the region bounded by the curves ( y = x^2 ) and ( y = 1 ) rotated about the line ( y = -2 ), you can use the method of cylindrical shells.

The integral to find the volume is given by:

[ V = \int_{a}^{b} 2\pi x (f(x) - g(x)) , dx ]

Where ( f(x) ) is the upper curve (in this case ( f(x) = 1 )), ( g(x) ) is the lower curve (in this case ( g(x) = x^2 )), and ( a ) and ( b ) are the x-values where the curves intersect.

First, find the intersection points of the curves by setting ( y = x^2 ) equal to ( y = 1 ):

[ x^2 = 1 ] [ x = \pm 1 ]

So, the bounds of integration are from ( x = -1 ) to ( x = 1 ).

Now, apply the formula for the volume:

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

[ V = 2\pi \int_{-1}^{1} (x - x^3) , dx ]

Integrate:

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

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

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

[ V = \frac{\pi}{2} ]

So, the volume of the solid generated by revolving the region about the line ( y = -2 ) is ( \frac{\pi}{2} ) cubic units.

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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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