## anonymous 4 years ago **Calc2 Help Needed** Find the surface area of the curve y= sqrt(x + 1) from x = 1 to x = 5 around the x-axis.

1. anonymous

I am getting 196/3 PI as my final answer.

2. anonymous

$2\pi \int\limits \sqrt(x+1) * \sqrt( 1 + (1/(2\sqrt(x+1)))^2$

3. anonymous

Someone help... Ahhh.. haha

4. TuringTest

I don't like it from there so I am about to do this in terms of y I am hopeful that will be prettier

5. TuringTest

yeah, definitely doable in terms of y.

6. TuringTest

$y=\sqrt{x+1}\to x=y^2-1$$\frac{dx}{dy}=2y$$ds=\sqrt{1+(2y)^2}dy$$S=2\pi\int_{\sqrt2}^{\sqrt6} yds=2\pi\int_{\sqrt2}^{\sqrt6}y\sqrt{1+4y^2}dy$$u=1+4y^2\to du=8ydy$so the integral is now$\frac{\pi}4\int\sqrt udu$

7. anonymous

which is 2/3 u^3/2..

8. TuringTest

...times pi/4 change back to y evaluate at limits

9. anonymous

Do i not need to change the limits since u-sub?

10. TuringTest

I would just change this back to terms of y before evaluating

11. TuringTest

2/3(1+4y^2)^(3/2) evaluated at sqrt2 to sqrt6

12. anonymous

so 3/2 x = y^3/2

13. anonymous

raise both sides to 2/3?

14. TuringTest

I'm not sure what you mean...

15. anonymous

would it be y = (3/2 *x)^2/3?

16. TuringTest

$y=\sqrt{x+1}\to x=y^2-1$$\frac{dx}{dy}=2y$$ds=\sqrt{1+(2y)^2}dy$$S=2\pi\int_{\sqrt2}^{\sqrt6} yds=2\pi\int_{\sqrt2}^{\sqrt6}y\sqrt{1+4y^2}dy$$u=1+4y^2\to du=8ydy$so the integral is now$\frac{\pi}4\int_{a}^{b}\sqrt udu=\frac{\pi}6u^{3/2}=\frac{\pi}6(1+4y^2)|_{\sqrt2}^{\sqrt6}$I don't see why you still have both y and x in what you are doing...

17. TuringTest

where do you get 3/2x ?

18. TuringTest

Anybody reading this if you could take a look at my solution here and find my error it would be greatly appreciated! http://openstudy.com/study#/updates/4f2b199ce4b039c5a5c85318

19. amistre64
20. amistre64

im assuming this is the same one as before ..

21. TuringTest

haha, I was trippin' earlier :D I could have done it just fine in x, but my brain wasn't working in the simplification

22. amistre64

$\int 2pi\ f(x)\sqrt{1+[f'(x)]^2}dxright?$

23. anonymous

integral of sqrt(4x + 5) is where I got to.

24. anonymous

Yep that is right.

25. TuringTest

hey amistre if you could have a look at the one in the link I posted that would be cool

26. amistre64

the link aint taking me anywhere

27. TuringTest

...by the way my answer is right as well, just for the record ;) http://www.wolframalpha.com/input/?i=integral%202pi%20y*sqrt(1%2B4y%5E2)dy%20from%20sqrt2%20to%20sqrt%206&t=crmtb01 @amistre then look at the last question asked by Cameron

28. anonymous

Oh wow 49/3 pi is right.. This thing is just really picky about decimals!

29. amistre64