i see, i got terms mixed up when transposing from a note book. so you agree the first answer is correct.

do you know why the method is incorrect, using trig functions instead?

i was a math major in college 20y ago, im relearning a lot of this just as a hobby, so i dont have a teacher/professor. the answer to my question may be apparent later as i go through the book. thanks for the help.

First, I apologize to you because I was wrong. Your set up for the first integral is correct. I have just realized that you have used \(\displaystyle 2y\) for one side of the square instead of \(\displaystyle 2x\) which is also correct.

To do this in trigonometry is very easy. You can follow skeeter way, or you can continue with my integral.

Last time, we had this integral,

\(\displaystyle V = \int_{-r}^{r} 4x^2 \ dy = 8\int_{0}^{r} r^2 - y^2 \ dy\)

Change the variables to polar coordinate.

\(\displaystyle y = r\sin\theta\)

\(\displaystyle dy = r\cos\theta \ d\theta\)

The integral will become,

\(\displaystyle V = 8\int_{0}^{\pi/2} (r^2 - r^2\sin^2 \theta) (r \cos \theta) \ d\theta = \frac{16r^3}{3}\)

Can you figure out why we chose the limit of integration as \(\displaystyle 0\leq \theta \leq \frac{\pi}{2}\)?