The region enclosed by the graph of y = x^2 , the line x = 2, and the x-axis is revolved abut the y-axis. The volume of the solid generated is:

A. 8pi
B. 32pi/5
C. 16pi/3
D. 4pi
5. 8pi/3

I solved for x as √y and set up this integral:

2pi * integral from 0 to 2 of y√y.

But it doesn't seem to give the answer, and I'm not sure what's wrong with it.

Using shells of thickness dx,

v = ∫[0,2] 2πrh dx
where r=x and h=y=x^2
v = ∫[0,2] 2πx^3 dx = 8π

using discs (washers) of thickness dy,

v = ∫[0,4] π(R^2-r^2) dy
where R=2 and r=x=√y
v = ∫[0,4] π(4-y) dy = 8π

You appear to have tried to rotate around the x-axis.

also, if you did that rotation, the height of the shells would be 2-x, not x.

Well, it looks like you've got the right idea with setting up an integral to find the volume of the solid generated by revolving the region. However, it seems like there might be a small error in your setup.

Instead of integrating y√y, you should be multiplying y by the circumference of the circle generated by revolving the given region. The formula for the circumference of a circle is 2πr, where r is the distance from the y-axis to the curve, which in this case is the value of x.

So, you would want to integrate 2πxy instead of just y√y. This would give you the correct expression for finding the volume of the solid generated.

To find the volume of the solid generated by revolving the region enclosed by the graph of y = x^2, the line x = 2, and the x-axis about the y-axis, you need to use the disk method.

First, let's find the limits of integration. The intersection points of y = x^2 and x = 2 are found by substituting x = 2 into y = x^2:

y = 2^2 = 4

So, the region of integration is y = 0 to y = 4.

To use the disk method, we need to express the radius (r) in terms of y. Since the region is being revolved about the y-axis, the radius is the distance from the axis of revolution to the curve at a given value of y.

The equation y = x^2 can be rewritten as x = √y. The distance from the y-axis to √y is √y. Therefore, the radius r is equal to √y.

The differential thickness of the disk is dy.

The volume of each individual disk is given by dV = π * r^2 * dy.

Substituting r = √y and integrating from y = 0 to y = 4, we get:

V = ∫[0 to 4] π * (√y)^2 * dy

V = ∫[0 to 4] π * y * dy

V = π * [y^2/2] [0 to 4]

V = π * (4^2/2 - 0^2/2)

V = π * (16/2)

V = 8π

Therefore, the volume of the solid generated by revolving the region enclosed by the graph of y = x^2, the line x = 2, and the x-axis about the y-axis is 8π.

So, the correct answer is A. 8π.

To find the volume of the solid generated by revolving the given region about the y-axis, you are correct in setting up your integral as:

V = 2π * ∫[from 0 to 4] (y * √y) dy

However, there seems to be a mistake in the limits of integration. The region enclosed by the graph of y = x^2, the line x = 2, and the x-axis is a quarter circle with a radius of 2 units. Therefore, when revolving this region about the y-axis, the limits of integration should correspond to the boundaries of the quarter circle.

To find these new limits, we need to find the y-values where the curve y = x^2 intersects the line x = 2. Setting x = 2 in the equation y = x^2, we have:

y = (2)^2
y = 4

So the upper limit of integration should be 4.

Now, let's set up the correct integral:

V = 2π * ∫[from 0 to 4] (y * √y) dy

Integrating this expression, we get:

V = 2π * (∫ y^(3/2) dy) (from 0 to 4)
= 2π * ((2/5) * y^(5/2)) (from 0 to 4)
= 2π * ((2/5) * (4^(5/2) - 0))

Simplifying further:

V = 2π * ((2/5) * (8 * √4))
= 2π * ((2/5) * (8 * 2))
= 2π * (32/5)
= 64π/5

Therefore, the correct answer is option B. 32π/5.