Refer to the figure and find the volume generated by rotating the given region about the specified line.
$ \Re_3 $ about $ AB $
Applications of Integration
in order to find the volume, you know, we integrate along the access parallel to the axis of rotation, which we know the excessive irritation is X equals one. Therefore, the outer radius is one minus y to the fourth cause it's the distance from X equals right of the fourth, the axis of rotation which we already establishes X equals one. Therefore, the inner radius is one minus y again the same reasons the distance from X equals y to the X invitation X equals one. Therefore, our volume is going to be pi times the integral from 01 of outer. Why minus one minus one of the fourth square minus inner one minus y squared. Do you? Why? So it's our money spinner. Just remember that formula. Before we integrate, I would recommend simplifying it. It makes things a lot easier to read when you integrate. If you have separate terms, you don't have things plugged in distributed form. Now that we've simplified, we can integrate using the power rule which means increased the exponent by one and divide by the new exponents. Now we're at the point where we can plug in, are bound zero and one when we know we have zero on the bottom bound, which gives a 17 pi over 45 is our solution to this problem.