Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified axis.

y = 10 x and y = 5 x^2 about y =0

Find the volume of the solid obtained by rotating the region bounded by y=8 x^2, x = 1, and y = 0 , about the x-axis.

I saw some other people were having trouble too and looked at those, but it makes no sense to me. Help? :(

ok. for volume generated by rotating around y=0, you can use discs or shells. The volume of a disc, or washer with a hole in it, is

v=π(R^2-r^2)h
where the height is dx and the radii are the distance from the x-axis to the two curves.

The upper curve is y=10x and the lower curve is y=5x^2. If you take a thin slice of the area and rotate it around the x-axis, you will get a washer.

Since the two curves intersect at (0,0) and (2,20), we integrate along x from 0 to 2.

v = ∫[0,2] π(R^2-r^2) dx
= π∫[0,2] (10x)^2 - (5x^2)^2 dx
= π∫[0,2] 100x^2 - 25x^4 dx
= π(100/3 x^3 - 5x^5) [0,2]
= π(800/3 - 160)
= 320/3 π

Now, you can also use shells. Recall that the volume of a thin cylinder of thickness dy is just

v = 2πrh dy

Now, in this case, the radius is just y, since we are rotating around the x-axis. The height h is the distance between the two curves

x = √(y/5) and
x = y/10

So, our shell volume is

v = ∫[0,20] 2πy(√(y/5)-y/10)
= 2π∫[0,20] 1/√5 y^(3/2) - y^2/10 dy
= 2π(2/5√5 y^(5/2) - y^3/30) [0,20]
= 2π(2/5√5 * 400√20 - 800/3)
= 2π(320 - 800/3)
= 320/3 π

Now see whether you can set up the other one. As you can probably see, using discs is easier in this case.

Thank you so much that makes sense more sense to me! :)

Certainly! I'll explain the process step by step to help you find the volume of the solids.

1. The first problem requires finding the volume of the solid obtained by rotating the region bounded by the curves y = 10x and y = 5x^2 about the y = 0 axis.

To solve this problem, we need to integrate the cross-sectional areas of the solid with respect to y.

a) First, let's find the points of intersection between the two curves: y = 10x and y = 5x^2.

Setting the equations equal to each other, we have:
10x = 5x^2

Rearranging the equation, we get:
5x^2 - 10x = 0

Factoring out 5x, we have:
5x(x - 2) = 0

So, x = 0 or x = 2.

b) Now, let's determine the limits of integration. Since we are rotating the region about the y = 0 axis, the interval of integration will range from y = 0 to the maximum y-value of the region.

Since both curves are positive for x values within the interval [0, 2], the maximum y-value of the region is given by y = 10x.

To find the maximum y-value, substitute x = 2 in y = 10x:
y = 10(2) = 20

So, the limits of integration are y = 0 to y = 20.

c) Next, let's express x in terms of y. For the curve y = 10x, rearranging, we have:
x = y/10.

For the curve y = 5x^2, rearranging, we get:
x = sqrt(y/5).

d) Now, we can calculate the cross-sectional area of the solid at an arbitrary value of y. The cross-sectional area of a cylindrical disk is given by A = πr^2, where r is the radius.

For this solid, the radius r is the distance between the two curves at a given value of y. It is given by:
r = x_upper - x_lower,

where x_upper represents the x-value on the curve y = 10x, and x_lower is the x-value on the curve y = 5x^2.

Substituting x = y/10 for x_upper and x = sqrt(y/5) for x_lower, we have:
r = y/10 - sqrt(y/5).

e) The cross-sectional area A at a specific value of y is then:
A = π(r^2) = π(y/10 - sqrt(y/5))^2.

f) Finally, we integrate the cross-sectional area A with respect to y, using the limits of integration determined earlier.

The volume of the solid can be calculated by integrating the cross-sectional area as follows:
V = ∫[0, 20] π(y/10 - sqrt(y/5))^2 dy.

Solving this definite integral will give you the volume of the solid.

Now, let's move on to the second problem.

2. The second problem asks for the volume of the solid obtained by rotating the region bounded by the curves y = 8x^2, x = 1, and y = 0 about the x-axis.

To solve this problem, we'll follow a similar process as before, but this time, we'll integrate the cross-sectional areas of the solid with respect to x.

a) First, let's determine the limits of integration. We are rotating the region about the x-axis, so the interval of integration will range from x = 0 to the maximum x-value of the region.

The maximum x-value is given by x = 1, which is the right endpoint of the interval.

So, the limits of integration are x = 0 to x = 1.

b) Next, let's express y in terms of x for the curve y = 8x^2. We don't need to rearrange the equation as it is already expressed in terms of x.

c) Now, we can calculate the cross-sectional area of the solid at an arbitrary value of x. The cross-sectional area is still given by A = πr^2, but this time, the radius r is the value of y.

So, the cross-sectional area A at a specific value of x is simply:
A = πy^2 = π(8x^2)^2 = π64x^4.

d) Finally, we integrate the cross-sectional area A with respect to x using the limits of integration determined earlier.

The volume of the solid can be calculated by integrating the cross-sectional area as follows:
V = ∫[0, 1] π64x^4 dx.

By solving this definite integral, you'll find the volume of the solid.

Remember, integrating the cross-sectional areas with respect to the appropriate variable will help you find the volume of solids obtained by rotating regions bounded by curves.