Set up, and evaluate the integral that represents the surface area of rotation of this cardioid about the y-axis.

[Elix]

I do not know if I set up the actual equation correctly so any tips would help! Thanks!

Cardioid:
[MATH]r=1+sin(theta)[/MATH]
Here is my work, I split it into smaller segments to keep it organized:

 

[MarkFL]

I don't see any errors in your work so far. Let's write (after some simplification):

[MATH]A=2\sqrt{2}\pi\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos(\theta)(\sin(\theta)+1)^{\frac{3}{2}}\,d\theta[/MATH]
Now it appears the integration should be straightforward.

 

[MarkFL]

To follow up, let's let:

[MATH]u=\sin(\theta)+1\implies du=\cos(\theta)\,d\theta[/MATH]
And our definite integral becomes:

[MATH]A=2\sqrt{2}\pi\int_0^2 u^{\frac{3}{2}}\,du=2\sqrt{2}\pi\left(\frac{2}{5}2^{\frac{5}{2}}\right)=\frac{32\pi}{5}[/MATH]

 

[Elix]

To follow up, let's let:

[MATH]u=\sin(\theta)+1\implies du=\cos(\theta)\,d\theta[/MATH]
And our definite integral becomes:

[MATH]A=2\sqrt{2}\pi\int_0^2 u^{\frac{3}{2}}\,du=2\sqrt{2}\pi\left(\frac{2}{5}2^{\frac{5}{2}}\right)=\frac{32\pi}{5}[/MATH]

Thanks a bunch! I struggled to connect how you got to that simplification at first but figured it out. I really appreciate your help with this and my previous question!