Calculus II

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Here is the picture I forgot earlier.

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The formula you need is

`S_x=2pi int_a^by (ds)/dx dx`

So in your case you have

`S_x=2piint_0^pi k(sint-tcost)(ds)/dx dx`

Since your curve is given parametrically you have `ds=sqrt((dx/dt)^2+(dy/dt)^2)dt`



since `cos^2t+sin^2t=1` we have  



Now we have two integrals for which we will use integration by parts




Similarly (we need to apply integration by parts two times) we obtain solution for the second integral.

`I_2=int t^2costdt=2tcost+(t^2-2)sint+C`

Now we combine those two integrals (add them)


and return back to calculation the surface area.



`S_x=6k^2pi^2` <--Your solution

Hope this helps.

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