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

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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 + 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[itex]\pi[/itex][(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[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! :)
 

Answers and Replies

  • #2
Dick
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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 + 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[itex]\pi[/itex][(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[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! :)
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).
 
  • #3
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=_=! 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!
 
  • #4
Dick
Science Advisor
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26,260
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=_=! 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.
 
  • #5
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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(

Ugh, please help me here. :/
I don't see how I can do a u substitution.
_____________________________________________________

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

= ∫2[itex]\pi[/itex][(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:
  • #6
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And it looks the 49,894 answer I got from Wolfram Alpha is wrong...
 
  • #7
Dick
Science Advisor
Homework Helper
26,260
619


And it looks the 49,894 answer I got from Wolfram Alpha is wrong...
No, it's not wrong. You just aren't following through with whole program. You were supposed to convert the integral to a du integral. If tan(u)=10x then sec^2(u)du=10dx. It's maybe somewhat easier if you substitute 10x=sinh(u) if you know hyperbolic trig functions. But it's not supposed to be a very easy problem.
 
Last edited:

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