STUPID ME. To be honest I looked at my notes and I was looking for where I got \sqrt{y^{2}-1} and I can't find it anywhere. Maybe I got it from my original integral which was in terms of x then I just removed the inside of the radical. It's things like that which I need to watch out for. Thanks...
I knew I was wrong there. So I solved for the values of y and I got √2 and √6 for my bounds in my second equation. Am I correct? OH! And thanks for noticing the 4y^2. I didn't notice that I did not square the Y
as for this one, yeah, I get what your trying to say. I changed my answer.
THIS...
Hi. I understand how to solve surface Area using integration when it is to be revolved about the x or y axis. But when the axis is not x or y I have a difficult time solving it. Please help me. Here is the equation
sqrt(x+1) rotated at x=-1 and y=5.
the bounds are 1 to 5.
since y=sqrt(x+1)...