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**"Find the area of the surface of the curve obtained by rotating the.."**

1. Find the area of the surface obtained by rotating the curve y= 1+5x^2

from x=0 to x=5 about the y-axis.

2. I thought to find surface area, we would need to use this formula:

SA= ∫2[itex]\pi[/itex](f(x))√(1 + (f'(x))

^{2})]dx

3. So far I have:

∫0 to 5 of 2[itex]\pi[/itex][(1 + 5x

^{2})(√(1+100x

^{2}))]dx

I first tried to do a u substitution for what's in the square root (1 + 100x

^{2}), but then du only equals

200x... And I'm not able to replace the 5x^2...

So it looks like u substitution will not work.

So then I thought I might be able to use the product rule.

So I did:

SA= ∫2[itex]\pi[/itex][(1+ 5x^2)(1/2(100x^2)

^{-1/2})*200x + ((1+100x

^{2})

^{1/2})(10x))]dx |0 to 5

But when I plug in 5 and 0 from here, I get a large number.

The number I get is 7521[itex]\pi[/itex].

But when I put this integral into Wolfram Alpha, it says the answer is an even larger number: 49,894!

I am confused as to what I am doing wrong!

Please help!

Thank you so much! :)