Use the method of cylindrical shells to find the volume V generated by rotating the region bounded by the?
Use the method of cylindrical shells to find the volume V generated by rotating the region bounded by the given curves about y = 8.
27y = x^3, y = 0, x = 6
V=
did you make a sketch?
did you notice that (6,8) is the farthest right-most point of your region?
So I will find the volume of the cylinder with radius 8 and height of 6, then subtract the region we don't want
cylinder = π(8^2)(6) = 384π
radius of region we don't want = 8 -(x^3/27)
volume of that rotated region
= π∫(8 - x^3/27)^2 dx from x = 0 to x = 6
= π∫(64 - (16/27)x^3 + x^6 /729) dx
= π [64x - (4/27)x^4 + (1/5103)x^7 ] from 0 to 6
= π(384 - 192 + 384/7 - 0 - 0 - 0)
= (1728/7)π
volume as asked for = 384π - 1728π/7 = 960π/7
better check my arithmetic, easy to make errors here without writing the solution down on paper.
Sure, let's find the volume using the method of cylindrical shells!
To use cylindrical shells, we need to integrate the circumference of the shell times its height.
First, let's find the limits of integration. The given curves are y = 0, x = 6, and 27y = x^3.
Since we want to rotate around y = 8, we need to find the intersection points between y = 8 and 27y = x^3.
Plugging in y = 8 into 27y = x^3 gives us 216 = x^3. Taking the cube root of both sides, we get x = 6.
Therefore, our limits of integration are from y = 0 to y = 8.
Now let's determine the radius and height of each cylindrical shell.
The radius of the cylindrical shell at any given y is given by the distance from the y-axis to the curve x = (27y)^(1/3). So the radius, r, is (27y)^(1/3).
The height of the cylindrical shell will be the distance between the curves x = 6 (vertical line) and x = (27y)^(1/3).
Therefore, the height, h, is 6 - (27y)^(1/3).
Now we can set up the integral to find the volume, V:
V = ∫[from y = 0 to y = 8] [(2πrh) dy]
= ∫[from y = 0 to y = 8] [(2π)(27y)^(1/3)(6 - (27y)^(1/3)) dy]
Since this integral involves calculating the cube root, it might get a bit complicated. So let's simplify things by introducing a new variable.
Let's substitute u = (27y)^(1/3), so du = (1/3)(27y)^(-2/3) * 27 dy.
The limits of integration change as well. When y = 0, u = 0, and when y = 8, u = 6.
Using the substitutions, our integral becomes:
V = ∫[from u = 0 to u = 6] [(2π)(u^3)(6 - u) * (3/27u^2) du]
= (2π/9) * ∫[from u = 0 to u = 6] (u^3)(6 - u) du
Now you can evaluate this integral to find the volume V. Good luck, and may your calculations be shellebratory!
To use the method of cylindrical shells to find the volume V, we need to integrate the product of the circumference of each cylindrical shell and its height over the interval defined by the given curves.
1. Find the intersection points of the given curves:
First, set the equations equal to each other to find the x-values at which they intersect:
27y = x^3 (equation 1)
y = 0 (equation 2)
Substitute equation 2 into equation 1:
27(0) = x^3
0 = x^3
x = 0
The only intersection point is (0, 0).
2. Determine the height of each cylindrical shell:
Since we are rotating about y = 8, the height of each cylindrical shell will be the difference between y = 8 and y = 0:
Height = 8 - 0 = 8.
3. Express the circumference of each cylindrical shell:
The circumference of each cylindrical shell at a given y-value is given by 2πr, where r is the distance from the y-axis to the cylindrical shell.
To find r, we need to solve equation 1 for x in terms of y:
27y = x^3
x^3 = 27y
x = (27y)^(1/3)
The distance from the y-axis to the cylindrical shell is simply the x-value, so r = (27y)^(1/3).
Thus, the circumference of each cylindrical shell is:
Circumference = 2πr = 2π(27y)^(1/3).
4. Set up the integral:
The volume V is given by the integral of the product of the circumference and the height over the interval of y-values from 0 to the y-value of intersection.
V = ∫[0 to 8] (2π(27y)^(1/3))(8 - 0) dy
5. Evaluate the integral:
V = 16π ∫[0 to 8] (27y)^(1/3) dy
We can simplify the integral by substituting u = (27y)^(1/3), and then du = (1/3)(27y)^(-2/3)dy.
Substituting u = (27y)^(1/3) and du = (1/3)(27y)^(-2/3)dy into the integral, we have:
V = 16π ∫[0 to 8] (27y)^(1/3) dy
V = 16π ∫[0 to 8] u du
Integrating u, we get:
V = 16π [u^2/2] [0 to 8]
V = 16π [(8^2/2) - (0^2/2)]
V = 16π (64/2)
V = 512π
Therefore, the volume generated by rotating the region bounded by the given curves about y = 8 is 512π cubic units.
To find the volume V generated by rotating the region bounded by the given curves about y = 8 using the method of cylindrical shells, we can follow these steps:
1. First, sketch the region bounded by the curves on a graph to visualize the shape of the region. In this case, the given curves are 27y = x^3, y = 0, and x = 6.
2. Determine the limits of integration. Since we are rotating about y = 8, we need to find the y-values where the curves intersect with y = 8. To do this, set 27y = x^3 equal to y = 8 and solve for x:
27(8) = x^3
216 = x^3
x = 6
So, the limits of integration for x are 0 to 6.
3. Next, write an expression for the height of each cylindrical shell. The height of each cylindrical shell is given by the difference between y = 8 and y = 0. So, the height function is h(y) = 8 - 0 = 8.
4. Determine an expression for the radius of each cylindrical shell. In this case, the radius is simply the x-coordinate of the curve at a given y-value. Since x = 6 is a constant value for the entire region, the radius function is r(y) = 6.
5. Find the differential volume of each cylindrical shell. The differential volume is given by the formula: dV = 2πrh(y)dy, where r is the radius and h(y) is the height.
6. Integrate the differential volume from the lower limit to the upper limit to find the total volume V. The integral expression is: V = ∫(0 to 6) 2π(6)(8)dy.
Simplifying the integral expression gives: V = 96π ∫(0 to 6) dy.
7. Evaluate the integral to find the volume V. The integral of dy is simply y. So, integrating the expression gives: V = 96π[ y ] (from 0 to 6).
Plugging in the limits of integration, we get: V = 96π[ 6 - 0 ] = 576π.
Therefore, the volume V generated by rotating the region about y = 8 is 576π cubic units.