Consider the region bounded by the graphs of the equations x=y^2 and y=3x. Set up 2 integrals, one with respect to x and the other with respect to y, both of which compute the volume of the solid obtained by rotating this region about the x-axis and evaluate the integrals.

Can you double check if there is a typo on the first function

x=y^2 => y=sqrt(x), ...(case 1) or is it
y=x^2 => x=sqrt(y).... (case 2)

In any case, the procedure is the same. I will work out the problem where y=x² (case 2).

Let f1(x)=3x (line), and
f2(x)=x² (curve)

Plot the two functions on a graph and determine the intersections.
Here is a plot (of case 2):
http://img854.imageshack.us/i/1299607814.png/

Call the intersections P1(0,0), P2(3,9).

For integration with respect to x, the limits of integration are x1,x2, and with respect to y, the limits are y1,y2.

1. integration with respect to x: (disk method).
Cut the region into vertical slices of thickness dx. When rotated about the x-axis, it will form a disk with a hole, of outside diameter f1(x), and inside diameter f2(x).
The volume of such a disk is therefore
dV=π(R²-r²)dx
=π(f1(x)²-f2(x)²)dx
The total volume is therefore
V=∫dV
=π∫(f1(x)²-f2(x)²)dx

2. Integrate with respect to y: (shell method)
Here, we need the inverse of each function, such that
for a given y=f1(x), x=g(y).
This can be obtained by solving for x in terms of y:
g1(y) = x = y/3, and
g2(y) = x = sqrt(y)

To find the volume, again we will be cutting slices, of thickness dy. However, when the slice is revolved around the x-axis, it becomes a shell (or hollow cylinder) of thickness dy, the height is (g2(y)-g1(y)), and the elemental volume
dV = surface area * thickness
= 2πrhdy
= 2π(y)(g2(y)-g1(y))dy
So the total volume
=∫dV
=2π∫y(sqrt(y)-y/3)dy
=2π∫(y^(3/2)-y²/3)dy
from y=0 to y=9

I get 162π/5 in both cases.

To set up the integrals, we need to determine the limits of integration for both x and y. Let's start by finding the points of intersection of the two curves.

First, let's solve the equations x = y^2 and y = 3x simultaneously. Substituting y = 3x into x = y^2, we get:

x = (3x)^2
x = 9x^2
9x^2 - x = 0

Factoring out x, we have:

x(9x - 1) = 0

This equation is satisfied when x = 0 or x = 1/9. Now let's find the corresponding y-values for these x-values.

For x = 0, using y = 3x, we have y = 3*0 = 0.

For x = 1/9, using y = 3x, we have y = 3*(1/9) = 1/3.

Therefore, the limits of integration for y will be from y = 0 to y = 1/3, and the limits of integration for x will be from x = 0 to x = 1/9.

To compute the volume of the solid obtained by rotating this region about the x-axis, we will set up two integrals.

1. Integral with respect to x:
We will use the method of cylindrical shells.

The radius (r) of each cylindrical shell can be calculated as the distance from the x-axis to the curve x = y^2. So, r = y^2.

The height (h) of each cylindrical shell can be calculated as the difference between the y-values of the curves y = 3x and y = 0. So, h = 3x - 0 = 3x.

The differential thickness (dx) will be used as usual.

Therefore, the integral with respect to x will be:

Vx = ∫ (2πrh) dx
= ∫ (2πy^2)(3x) dx
= 6π ∫ x * y^2 dx

Substituting x = y^2, the integral becomes:
Vx = 6π ∫ (y^2)*(y^2) dy
= 6π ∫ y^4 dy
= 6π * (1/5) * [y^5] between y = 0 and y = 1/3

Evaluating this integral, we have:
Vx = 6π * (1/5) * [(1/3)^5 - 0]
= 6π * (1/5) * (1/243)
= 6π/1215

2. Integral with respect to y:
We will use the method of washers.

The outer radius (R) of each washer can be calculated as the distance from the x-axis to the curve x = y^2. So, R = y^2.

The inner radius (r) of each washer can be calculated as the distance from the x-axis to the line y = 0. So, r = 0.

The height (h) of each washer can be calculated as the difference between the x-values of the curves x = y^2 and x = (1/9). So, h = y^2 - (1/9).

The differential thickness (dy) will be used as usual.

Therefore, the integral with respect to y will be:

Vy = ∫ (π(R^2 - r^2)h) dy
= ∫ π(y^2 - 0^2)(y^2 - 1/9) dy
= π ∫ (y^2)(y^2 - 1/9) dy

Simplifying this expression, we have:
Vy = π ∫ (y^4 - 1/9 * y^2) dy
= π * (1/5 * [y^5] - 1/9 * [y^3]) between y = 0 and y = 1/3

Evaluating this integral, we have:
Vy = π * (1/5 * (1/3)^5 - 1/9 * (1/3)^3)
= π/1215

Therefore, the volume of the solid obtained by rotating this region about the x-axis is:

V = Vx + Vy
= 6π/1215 + π/1215
= 7π/1215

Thus, the volume of the solid is 7π/1215.

To find the volume of the solid obtained by rotating the region bounded by the graphs of the equations x = y^2 and y = 3x about the x-axis, we can use the method of cylindrical shells.

First, let's find the points of intersection between the two curves by setting them equal to each other:

y^2 = 3x

To solve for x, we can square both sides:

x = (1/3)y^4

Now we have the equations in terms of y. To determine the limits of integration, we need to find the y-values at which the curves intersect. We can do this by setting the equations equal to each other:

(1/3)y^4 = 3y

Divide both sides by y:

(1/3)y^3 = 3

Now solve for y:

y^3 = 9

y = ∛9

The limits of integration for y are from 0 to ∛9 since the curves intersect at y = ∛9.

Now let's set up the integral with respect to y to compute the volume:

V = ∫[0,∛9] 2πy * h(y) * dy

where h(y) is the height of the cylindrical shell and is equal to the difference between the x-values at a given y.

From the equation x = y^2, we can solve for x:

x = (1/3)y^4

The height, h(y), is then:

h(y) = (1/3)y^4 - y^2

Substituting this into the integral, we get:

V = ∫[0,∛9] 2πy * ((1/3)y^4 - y^2) dy

Integrating this expression will give us the volume of the solid.

Now, we can set up the integral with respect to x:

V = ∫[0,3] 2πx * h(x) * dx

Similarly, we need to find the y-values at which the curves intersect for the limits of integration. Setting the equations x = y^2 and y = 3x equal to each other, we have:

y^2 = 3y

Solving for y:

y^2 - 3y = 0

y(y - 3) = 0

So, y = 0 or y = 3. The limits of integration for x are from 0 to 3 since the curves intersect at x = 3.

Now, we can find the height, h(x), by solving for y in terms of x from the equation y = 3x:

x = y/3

y = 3x

Substituting this into the equation x = y^2:

x = (3x)^2

x = 9x^2

1 = 9x

x = 1/9

Therefore, the height, h(x), is:

h(x) = (3x) - (1/9)

Substituting this into the integral, we have:

V = ∫[0,3] 2πx * ((3x) - (1/9)) dx

Integrating this expression will give us the volume of the solid.

Note: In order to evaluate the integrals and obtain the final volume, numerical methods or advanced integration techniques may be required.