Hello everyone, I was wondering if someone could check my work on this problem as I'm not sure it's right. I'm in Calc II right now and we are doing finding volume bounded by curves rotated around an axis.
So, here is the problem, y=cos(pix)+1, y=4x^(2)-9, x=0; about x=0
I wasn't sure if the bounds were to be 0 to the intersection point of 1.607006, or from -1.607006 to 1.607006. But I did the integration work shown below.
I used the shell method.
2pi * the integral of (x)(cos(pix)+1) - (4x^(2)-1) (this is for the height)
2pi* the integral of (xcos(pix)-4x^(3)+10x)
I used integration by parts for the xcos(pix) and got u=x, du=dx, dv=cos(pix), v=1/pi*sin(pix)
I went through doing the parts and I came up with x(1/pi*sin(pix)) - integral of VDU
I did the integral, did the integrals of 4x^(3) and 10x, put everything together and came up with
2pi* [(x)(1/pi*sin(pix)) + 1/(pi)^(2) * cos(pix) -x^(4) +5x^(2)]+C
Is this correct??? Thank you very very much for whoever looks over this, I know this is a lot of work.