Yes, as mmm said, it is just trig.
Look at the little red triangle with sides du and ds. \(\displaystyle \frac{\text{adjacent}}{\text{hypoteneuse}}=\frac{du}{ds}=cos(\beta-\alpha)\)
See?. As mmm stated, you can think of du, ds, and so on as sides to a triangle.
Here is a graph. I tried to label everything. By the way, did you get part 4?. Rotating x+sin(x) about y=x-2 from 0 to 2Pi
Look at the little red triangle with sides du and ds. \(\displaystyle \frac{\text{adjacent}}{\text{hypoteneuse}}=\frac{du}{ds}=cos(\beta-\alpha)\)
See?. As mmm stated, you can think of du, ds, and so on as sides to a triangle.
Here is a graph. I tried to label everything. By the way, did you get part 4?. Rotating x+sin(x) about y=x-2 from 0 to 2Pi
