Find the volume of the solid obtained by rotating the region enclosed by the graphs of y=18-x, y=3x-6 and x=0 about the y-axis

V=

To find the volume of the solid obtained by rotating the region enclosed by the given graphs about the y-axis, we'll use the method of cylindrical shells.

Step 1: Determine the limits of integration for y.
First, let's find the points of intersection between the curves. Setting y=18-x and y=3x-6 equal to each other:
18-x = 3x-6
4x = 24
x = 6

So the curves intersect at x=6. Now let's find the y-values of these intersection points using the equation y=18-x:
y = 18-6
y = 12

Therefore, the limits of integration for y are from y=12 to y=0.

Step 2: Set up the integral for the volume.
The volume of a cylindrical shell is given by the formula V = ∫2πx*f(x) dx, where f(x) is the height of the shell at a given x-value and x is the distance from the axis of rotation.

Since we are rotating about the y-axis, the distance from the axis of rotation is x. And the height of the shell can be calculated as the difference between the upper curve and the lower curve at each x-value.

So, the integral for the volume V is:
V = ∫[from y=12 to y=0] 2πx * [ (18-x) - (3x-6) ] dy

Step 3: Evaluate the integral.
V = 2π ∫[from y=12 to y=0] x (12 + 6 - 3x) dy
V = 2π ∫[from y=12 to y=0] x (18 - 3x) dy

Now we need to express everything in terms of y, since that is our variable of integration. Rearranging the equation y=18-x, we get x=18-y, and substituting this into the integral, we have:
V = 2π ∫[from y=12 to y=0] (18 - 3(18-y)) y dy
V = 2π ∫[from y=12 to y=0] (18 - 54 + 3y) y dy
V = 2π ∫[from y=12 to y=0] (3y^2 + 18y - 54) dy

Now we can integrate:
V = 2π [ y^3/3 + 9y^2/2 - 54y ] [from y=12 to y=0]
V = 2π [ (12^3/3 + 9(12)^2/2 - 54(12)) - (0^3/3 + 9(0)^2/2 - 54(0)) ]

Calculating this expression will give us the final volume value.

To find the volume of the solid obtained by rotating the region enclosed by the given graphs about the y-axis, you can use the method of cylindrical shells.

1. First, let's determine the limits of integration. We need to find the y-values where the graphs intersect. Setting y=18-x and y=3x-6 equal to each other, we can solve for x:

18 - x = 3x - 6
4x = 24
x = 6

So, the graphs intersect at x=6.

2. Next, let's set up the integral. The general formula for the volume using cylindrical shells is:

V = ∫(2πrh)dy

where:
- r is the radius of each shell, which is the distance from the y-axis to the function curve
- h is the height of each shell, which is the infinitesimal change in y
- dy is the infinitesimal thickness of each shell

Since we are rotating about the y-axis, the distance from the y-axis to the function curve can be expressed as x. So, r = x.

The height of each shell, h, is simply dy.

3. Now, let's express the integral in terms of y. The region enclosed by the given graphs has two parts: below and above x=6.

- For y-values where 0 ≤ y ≤ 12 (below x=6):
The range of y for this part is from 0 to 12.
For each y-value in this range, the corresponding values of x are given by the function equations.
So, we can express the integral as:
V1 = ∫(0 to 12) [2π(x)(dy)], where the limits of integration are for y.

- For y-values where 12 ≤ y ≤ 18 (above x=6):
The range of y for this part is from 12 to 18.
For each y-value in this range, the corresponding values of x are given by the other function equations.
So, we can express the integral as:
V2 = ∫(12 to 18) [2π(x)(dy)], where the limits of integration are for y.

4. Finally, let's evaluate the integrals and find the total volume:
V = V1 + V2 = ∫(0 to 12) [2π(x)(dy)] + ∫(12 to 18) [2π(x)(dy)]

You can evaluate these integrals using the limits of integration and the equations for x as found earlier. Simplify the expression and evaluate the definite integrals to find the volume of the solid.