Let R be the region bounded by y=x^2, x=1, and y=0. Use the shell method to find the volume of the solid generated when R is revolved about the line x=-11.
To find the volume of the solid generated when region R is revolved about the line x = -11 using the shell method, follow these steps:
1. Sketch the region R on a coordinate plane. R is bounded by the curve y = x^2, the vertical line x = 1, and the x-axis.
2. Draw a vertical line at x = -11 to represent the axis of rotation.
3. Note that the region R lies to the right of the axis of rotation. To apply the shell method, we need to express the region in terms of x.
4. Notice that y = x^2 intersects the x-axis at x = 0 and x = 1. However, only the portion of the curve where x is greater than or equal to 1 contributes to the volume when revolved around x = -11.
5. To express the region R in terms of x, we need to find the upper and lower limits of integration. The lower limit is x = 1, which represents the vertical line bounding the region R. The upper limit can be determined using the equation of the vertical line x = 1, as it intersects the curve y = x^2.
6. To find the intersection point, set y = x^2 equal to y = 0 (the x-axis) and solve for x: x^2 = 0. This implies that x = 0.
7. Thus, the limits of integration for x are from 1 to 0.
8. Now, we can set up the integral using the shell method formula:
V = 2π ∫ (x)(height)(width) dx,
where the height is the distance traveled by the shell as it rotates around x = -11 and the width is the differential thickness of the shell along the x-axis.
9. Since we are rotating about the line x = -11, the height of each shell will be -11 - x.
10. The width of each shell is dx, which represents an infinitesimally small change in x.
11. Substitute these values into the integral:
V = 2π ∫ (-11 - x)(x)(dx)
12. Simplify the expression inside the integral:
V = 2π ∫ (-11x - x^2)(dx)
13. Integrate the expression using the limits of integration from 1 to 0:
V = 2π ∫ [-11x - x^2] dx, from 1 to 0.
14. Evaluate the integral:
V = 2π [-11 * (0 - 1) - (0^2 - 1^2)]
V = 2π [-11 * (-1) - (0 - 1)]
V = 2π [-11 + 1]
V = 2π (-10)
V = -20π
15. The volume of the solid generated when region R is revolved about the line x = -11 using the shell method is -20π cubic units.