Finding Surface Area

[Chris686]

I'm not sure how to go forward with this problem, as I can only find examples that still have x values underneath the radical. Here it is:

y=7x, 0<=x<=1

Formula = 2pi*f(x)*sqrt(1+(f'(x))^2)dx

So

From 0 to 1
2pi*7x*sqrt(1+7(^2))dx
2pi*7x*sqrt(50)dx
14pi*x*sqrt(25*2)dx
70pi*x*sqrt(2)dx

Now, from here (assuming I did everything correctly), where do I go for integrating?

Are my u values now from 0 to sqrt(2)?
If so, do I integrate the entire value, x*sqrt(2)?

 

[HallsofIvy]

Surface area of what? You don't mention any surface to have an area! From what you are doing, it suspect that you mean the surface formed by rotating the line y= 7x, x from 0 to 1, around the x axis. You arrive at \(\displaystyle \int_0^1 70\pi x\sqrt{2}dx\). You recognize, don't you, that \(\displaystyle 70\pi\sqrt{2}\) is a constant and can be taken outside the integral? Can you integrate \(\displaystyle 70\pi\sqrt{2}\int_0^1 x dx\)?

 

[Chris686]

Yes, that's it. Just copied it from the book.

Do'h! Now it's so obvious! Thanks!