how do you set up the volume of the region bounded by y=x and y=x^2 rotated about the line y=3.
To set up the volume of the region bounded by y = x and y = x^2, rotated about the line y = 3, you can use the method of cylindrical shells. Here's how you can do it step by step:
1. First, graph the two functions y = x and y = x^2 to visualize the region of interest. The region is bounded by the curves y = x and y = x^2.
2. Determine the limits of integration. Since we are rotating the region about the line y = 3, we need to find the x-values where the two curves intersect. Set the two equations equal to each other and solve for x:
x = x^2 ⟹ x^2 - x = 0.
Factoring out x, we get:
x(x - 1) = 0.
This gives us two x-values: x = 0 and x = 1.
Therefore, the limits of integration will be from x = 0 to x = 1.
3. Set up the integral for the volume using the method of cylindrical shells. The volume of each shell is given by the formula:
dV = 2πrh(x)dx,
where r(x) is the radius of each shell and h(x) is the height of each shell.
In this case, the radius r(x) will be the distance between the line y = 3 and the curve y = x (or y = x^2). So, the radius is: r(x) = 3 - x (for y = x), and r(x) = 3 - x^2 (for y = x^2).
The height h(x) will be the difference in y-coordinates between the two curves: h(x) = x - x^2.
Therefore, the integral to find the volume will be:
V = ∫[0 to 1] 2π [(3 - x)(x - x^2)] dx.
4. Evaluate the integral. Perform the integration of the expression using the limits of integration from step 2:
V = ∫[0 to 1] 2π [(3x - x^2) - (x^2 - x^3)] dx.
Simplifying the expression gives:
V = ∫[0 to 1] 2π (4x^3 - 3x^2) dx.
Integrate term by term to get:
V = 2π [(x^4/2) - (x^3)] evaluated from x = 0 to x = 1.
Plugging in the limits of integration and simplifying further gives:
V = 2π [(1/2) - 1] = -π/2.
Note that the negative volume indicates that the region is below the line y = 3.
Therefore, the volume of the region bounded by y = x and y = x^2, when rotated about y = 3, is -π/2 cubic units.