# Find the area of the surface of the curve obtained by rotating the

"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$\pi$(f(x))√(1 + (f'(x))2)]dx

3. So far I have:

∫0 to 5 of 2$\pi$[(1 + 5x2)(√(1+100x2))]dx

I first tried to do a u substitution for what's in the square root (1 + 100x2), 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$\pi$[(1+ 5x^2)(1/2(100x^2)-1/2)*200x + ((1+100x2)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$\pi$.

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!
Thank you so much! :)

Related Calculus and Beyond Homework Help News on Phys.org
Dick
Homework Helper

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$\pi$(f(x))√(1 + (f'(x))2)]dx

3. So far I have:

∫0 to 5 of 2$\pi$[(1 + 5x2)(√(1+100x2))]dx

I first tried to do a u substitution for what's in the square root (1 + 100x2), 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$\pi$[(1+ 5x^2)(1/2(100x^2)-1/2)*200x + ((1+100x2)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$\pi$.

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!
Thank you so much! :)
There is no real product rule for integrals, I'm not sure what you are thinking there. But to do the integration you need a trig substitution. Like 10x=tan(t).

=_=! Oh wow, I don't know what I'm thinking either! Of course product rule can't be used for integrals! Ugh. wow.

Okay, so you say that a trig substitution has to be used.
But how can we replace 10x with tan(x)?

So do you mean that in (√1 + (10x)2), we replace the 10x with tan(x)?

...And how would that help us?
Thanks!

Dick
Homework Helper

=_=! Oh wow, I don't know what I'm thinking either! Of course product rule can't be used for integrals! Ugh. wow.

Okay, so you say that a trig substitution has to be used.
But how can we replace 10x with tan(x)?

So do you mean that in (√1 + (10x)2), we replace the 10x with tan(x)?

...And how would that help us?
Thanks!
You don't just replace 10x with tan(x)! This is u-substitution, tan(u)=10x. The sqrt(1+100x^2)=sqrt(1+tan(u)^2)=sqrt(sec(u)^2)=sec(u). And, of course, you also need to figure out how du is related to dx. The usual substitution routine.

Hmmmm.

I think I see what you mean now about the tan(u)=10x for the u substitution.

tan(u)=10x

u= arctan(10x)

du=(1/((10x)2 + 1)*10dx

du= 10/(100x2 + 1)dx

I need to make the dx side have an x^2!

(100x2 + 1)du = 10dx

(1 + (1/100x2)du = (10/100x2)dx

50x4*(1 + (1/100x2))du = (1/10x2)dx *50x4

(50x4 + 1/2x2)du = 5x2dx

Well, I got a 5x2 on one side, but this does not help me at all because I still have x's on the other side! X(

I don't see how I can do a u substitution.
_____________________________________________________

∫0 to 5 of 2$\pi$[(1 + 5x2)(√(1 + (10x)2))]dx

= ∫2$\pi$[(1 + 5x2)(√(1 + (tan(u))2))]dx

So, I haven't finished my u substitution because I cannot get it to work...
Thanks for the help!

Last edited:

And it looks the 49,894 answer I got from Wolfram Alpha is wrong...

Dick