find the volume generated by revolving the area bounded by y=-x^3, y=0, and x=-2 about the line x=1.
I know how to the solve the problem but am not sure how to get the interval. Please help.
Well, if you graph things, you have a curved triangular region with vertices at (-2,8),(-2,0),(0,0).
So, using shells of thickness dx, you have the area as
∫[-2,0] 2πrh dx
where
r = 1-x
h = y = -x^3
using washers of thickness dy, you have
∫[0,8] π(R^2-r^2) dy
where
R = 1+2
r = 1+x = 1+∛y
To find the interval for which you need to revolve the area bounded by the curves around the line x=1, you can start by graphing the curves y=-x^3 and y=0.
The graph of y=0 is simply the x-axis.
The graph of y=-x^3 is a downward-opening cubic curve. When you graph it, you'll see that it intersects the x-axis at x=-1, x=0, and x=1.
Since the line x=1 is the axis of rotation, the interval for which you need to revolve the area is determined by the intersection points of the curves and the line x=1.
In this case, the interval is from x=-1 to x=1.
So, the interval for revolving the area bounded by the curves y=-x^3, y=0, and x=-2 about the line x=1 is from x=-1 to x=1.
Now, you can proceed to calculate the volume generated using your preferred method, such as using the disk or washer method.