Question

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

Answer 1

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

Answer 2

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

Answer 3

Someone help... Ahhh.. haha

Answer 4

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

Answer 5

yeah, definitely doable in terms of y.

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

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

Answer 12

so 3/2 x = y^3/2

Answer 13

raise both sides to 2/3?

Answer 14

I'm not sure what you mean...

Answer 15

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

Answer 16

\[y=\sqrt{x+1}\to x=y^2-1\]\[\frac{dx}{dy}=2y\]\[ds=\sqrt{1+(2y)^2}dy\]\[S=2\pi\int_{\sqrt2}^{\sqrt6} yds=2\pi\int_{\sqrt2}^{\sqrt6}y\sqrt{1+4y^2}dy\]\[u=1+4y^2\to du=8ydy\]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...

Answer 17

where do you get 3/2x ?

Answer 18

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

Answer 19

http://www.wolframalpha.com/input/?i=integrate+pi%28sqrt%284x%2B5%29%29+from+1+to+5

Answer 20

im assuming this is the same one as before ..

Answer 21

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

Answer 22

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

Answer 23

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

Answer 24

Yep that is right.

Answer 25

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

Answer 26

the link aint taking me anywhere

Answer 27

...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

Answer 28

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

Answer 29

just one of those days ..... still cant get a bead on your link

Answer 30

http://openstudy.com/users/cameronmx9#/updates/4f2aaff5e4b049df4e9e3d87