# 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?

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

the radius of the larger washer is the greater

the radius of the smaller washer is the lesser

The representative slice has volume

We need to evaluate the integral

# = 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# .

Theradiusof the representative shell is#y+2# (dotted line in picture)

Theheightgoes 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#

By signing up, you agree to our Terms of Service and Privacy Policy

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.

By signing up, you agree to our Terms of Service and Privacy Policy

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.

- How do you find the average value of the function for #f(x)=2xsqrt(1+x^2), -3<=x<=3#?
- What is the arc length of #f(x)=x^2/(4-x^2) # on #x in [-1,1]#?
- How do you use the method of cylindrical shells to find the volume of the solid obtained by rotating the region bounded by #y = 2 - (x^2)# and #y = x^2# revolved about the x=1?
- What is the general solution of the differential equation #(x+y)dx-xdy = 0#?
- How do you solve #y'=-xy+sqrty# given y(0)=1?

- 98% accuracy study help
- Covers math, physics, chemistry, biology, and more
- Step-by-step, in-depth guides
- Readily available 24/7