Find the volume of the solid obtained by revolving the region x=(y-2)^2, the x-axis, the y-axis, about the x-axis.

So I understand that you're supposed to use disc method; however, idk where I evaluated wrong. My final answer is 136π/3, but according to Wolfram Alpha, the answer is 8π/3 and they used shell. Below is my work:

x=(0-2)^2
x=(-2)^2
x=4

x=(y-2)^2
x^(1/2)=y-2
x^(1/2)+2=y

π∫(x^(1/2)+2)^2 dx [0,4]
π∫(x+4(x)^(1/2)+4) dx [0,4]
π[(x^2/2)+(8/3)x^(3/2)+4x] [0,4]
=136π/3

using discs of thickness dx,

v = ∫[0,4] πr^2 dx
where r = y = 2-√x
The parabola has two branches, and you took the upper branch 2+√x, but we want the curve bounded by the axes, which is 2-√x
v = ∫[0,4] π(2-√x)^2 dx

Make your sketch.

x = (y-2)^2
±√x = y - 2
y = 2 + √x or y = 2 - √x
the region that you are revolving has as its boundary y = 2 - √x
and crosses the x-axis at (4,0)
so we want:
π∫ y^2 dx from x=0 to x=4
= π∫ (2-√x)^2 dx from 0 to 4 = π∫ (2-x^(1/2))^2 dx from 0 to 4
= π∫ (4 - 4x^(1/2) + x) dx from 0 to 4
= π [4x - (8/3)x^(3/2) + (1/2)x^2 ] from 0 to 4
= π( 8 -(8/3)(8) + 8 - 0 )
= 8π/3 , as required

To find the volume of the solid obtained by revolving the region x = (y - 2)^2, the x-axis, the y-axis about the x-axis, we can use the method of disks or shells.

Using the disk method:
To evaluate the integral and determine where you went wrong, let's first set up the integral using the disk method.

The formula for the volume using the disk method is V = π ∫(R(x))^2 dx, where R(x) is the radius of a disk.

In this case, since we are rotating about the x-axis, the radius of the disk is given by R(x) = y = (x + 2)^(1/2), where x ranges from 0 to 4.

So the integral becomes:
V = π ∫[(x + 2)^(1/2)]^2 dx [0, 4]
V = π ∫(x + 2) dx [0, 4]
V = π [(x^2/2) + 2x] [0, 4]
V = π [(4^2/2) + 2(4)] - π [(0^2/2) + 2(0)]
V = π [8 + 8] - π [0 + 0]
V = 16π

Therefore, the volume of the solid is 16π.

It seems you may have made a mistake in the integration step, resulting in the incorrect answer of 136π/3. The correct answer using the disk method is 16π.

To find the volume of the solid obtained by revolving the region x = (y - 2)^2, the x-axis, the y-axis about the x-axis, you can use either the disk method or the shell method.

Since you mentioned that you used the disk method, let's analyze your steps to see where the mistake may have occurred.

1. The function x = (y - 2)^2 represents the right side of the region that needs to be revolved. In other words, it's the top boundary of the region.

2. To find the volume using the disk method, you need to consider an infinitesimally thin, cylindrical disk with thickness dx at a distance x from the axis of revolution.

3. The radius of this disk is given by the distance between the axis of revolution (the x-axis) and the function x = (y - 2)^2.

4. The volume of each disk can be calculated as π(radius^2)dx.

Now, let's proceed with the correct steps using the disk method:

1. Find the limits of integration. The region is bounded by x = 0 and x = 4 (as you correctly determined).

2. Express the radius of each disk in terms of x. The distance between the x-axis and the function x = (y - 2)^2 is simply y. Rearranging y = x^(1/2) + 2, we can obtain y = (x^(1/2)) + 2, which is the radius of each disk.

3. Square the radius to get r^2 = (x^(1/2) + 2)^2.

4. Integrate the volume of each disk with respect to x:

V = ∫ [π(r^2)dx] from x = 0 to x = 4
V = π ∫ [(x^(1/2) + 2)^2]dx from x = 0 to x = 4

Expanding and integrating, we have:

V = π ∫ [(x + 4(x)^(1/2) + 4)] dx from x = 0 to x = 4

Simplifying further, we get:

V = π [(x^2/2) + (8/3)x^(3/2) + 4x] from x = 0 to x = 4

5. Evaluate the definite integral:

V = π [((4^2)/2) + (8/3)(4^(3/2)) + 4(4)] - π [((0^2)/2) + (8/3)(0^(3/2)) + 4(0)]
V = π [8 + 128/3 + 16] - π [0 + 0 + 0]
V = π [136/3]
V = 136π/3

Based on the correct calculation, your initial calculation of the volume as 136π/3 is actually correct. The answer you obtained matches your earlier solution.

It seems that there might be an error in comparing your answer with Wolfram Alpha's answer, as both calculations should yield the same result using the same method.

If Wolfram Alpha used the shell method and obtained a different result, it might indicate that the two methods yield different volume formulas for this particular problem.