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Finding Surface Area of a Solid of Revolution

  • Thread starter BraedenP
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  • #1
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Homework Statement


Find the surface area generated by rotating [itex]y=5-4x^(3/2), 0\leq x\leq 1[/itex] about x=2.


Homework Equations



[tex]SA = 2\pi\int_{a}^{b}(r\cdot ds)dx[/tex]

The Attempt at a Solution



I simply filled in the formula for the given question, and I'm getting stuck at integration time.

[tex]SA = 2\pi\int_{0}^{1}(2-x)\cdot \sqrt{1+\left(\frac{dy}{dx}5-4x^{3/2})\right)^2}[/tex]

Simplified, it's:

[tex]SA = 2\pi\int_{0}^{1}(2-x)\cdot \sqrt{1+36x}[/tex]

We haven't done Integration By Parts yet, so I can't "deal with" the multiplication. What I tried to do was square the (2-x) to get [itex]x^2-4x+4[/itex] and then combined the roots:

[tex]SA = 2\pi\int_{0}^{1}\sqrt{36x^3-143x^2+140x+4}[/tex]

But that still isn't a manageable integral, given that I can't use substitution or anything.

How would I go about solving this question?
 
Last edited:

Answers and Replies

  • #2
rock.freak667
Homework Helper
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[tex]
SA = 2\pi\int_{0}^{1}(2-x)\cdot \sqrt{1+36x}
[/tex]


Try something simpler like a substitution of t=1+36x (remember, you can always rewrite x in terms of t :wink:)
 
  • #3
33,496
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Just to keep the account books straight, this integral
[tex]SA = 2\pi\int_{a}^{b}(r\cdot ds)dx[/tex]

should be
[tex]SA = 2\pi\int_{a}^{b}r\cdot ds[/tex]

Multiplying by dx/dx gives
[tex]SA = 2\pi\int_{a}^{b}r\cdot ds/dx~dx[/tex]

Using your function, we get
[tex]SA = 2\pi\int_{0}^{2}(2-x)\cdot \sqrt{1+36x}dx[/tex]


Notice the change in the upper limit of integration.
 
  • #4
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Gah, I always write the question wrong on here for some reason. I should have written [itex]0 \leq x \leq 1[/itex] above. (Changed now for clarity)

Basically, I'm fine getting that formula, but I get stuck when trying to integrate it. It's not in a format that I'm used to working with.
 
  • #5
33,496
5,188
I would go with rockfreak's suggested substitution.
 
  • #6
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Hmm, not quite sure why I can't get this question... Here's my attempt with the substitution:

[tex]SA = 2\pi\int_{0}^{1}(2-\frac{1}{36}(t-1))\cdot \sqrt{t}[/tex]

Expand everything:

[tex]SA = 2\pi\int_{0}^{1}2\sqrt{t}-\frac{1}{36}(t^{\frac{3}{2}}-\sqrt{t})[/tex]

Simplify:

[tex]SA = \frac{\pi}{18}\int_{0}^{1}72t^\frac{1}{2}-t^\frac{3}{2}+t^\frac{1}{2}[/tex]

And I get (we don't need to find a decimal answer -- integrating it to this form should be enough:

[tex]SA = \frac{\pi}{18}\left[ 48t^\frac{3}{2}-\frac{2}{5}t^\frac{5}{2}+\frac{2}{3}t^\frac{3}{2} \right ][/tex] from t=1 to t=37

I don't think this is right though...
 
  • #7
33,496
5,188
You keep omitting the dx or dt. That's the reason for a mistake you make right near the beginning. Your substitution was t = 1 + 36x, so dt = 36dx.

Also, when you are integrating, it's almost always better to write radical expressions using exponents.

Hmm, not quite sure why I can't get this question... Here's my attempt with the substitution:

[tex]SA = 2\pi\int_{0}^{1}(2-\frac{1}{36}(t-1))\cdot \sqrt{t}[/tex]

Expand everything:

[tex]SA = 2\pi\int_{0}^{1}2\sqrt{t}-\frac{1}{36}(t^{\frac{3}{2}}-\sqrt{t})[/tex]

Simplify:

[tex]SA = \frac{\pi}{18}\int_{0}^{1}72t^\frac{1}{2}-t^\frac{3}{2}+t^\frac{1}{2}[/tex]

And I get (we don't need to find a decimal answer -- integrating it to this form should be enough:

[tex]SA = \frac{\pi}{18}\left[ 48t^\frac{3}{2}-\frac{2}{5}t^\frac{5}{2}+\frac{2}{3}t^\frac{3}{2} \right ][/tex] from t=1 to t=37

I don't think this is right though...
 
  • #8
96
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Hmm, thanks.. Not sure why I omitted it. I did keep in mind while answering the question, however, which variable I was integrating with respect to.

Did my omission cause a mistake in the final answer, or were you just pointing out that I forgot to write it?
 
Last edited:
  • #9
33,496
5,188
It caused a mistake.
[tex]SA = 2\pi\int_{0}^{1}(2-x)\cdot \sqrt{1+36x}dx[/tex]

Let u = 1 + 36x, so du = 36dx, and x = (1/36)(u - 1). (I'm more used to using u for substitutions.)

With the substitutions, we have
[tex]SA = \pi/18 \int_{u = 1}^{37} [2 - (1/36)(u - 1)]u^{1/2}du[/tex]
[tex]= \pi/18 \int_{u = 1}^{37} [2u^{1/2} - (1/36)u^{3/2} + (1/36)u^{1/2}]du[/tex]

By omitting dx, you neglected a factor of 1/36 with du (since dx = (1/36)du).
 
  • #10
96
0
It caused a mistake.
[tex]SA = 2\pi\int_{0}^{1}(2-x)\cdot \sqrt{1+36x}dx[/tex]

Let u = 1 + 36x, so du = 36dx, and x = (1/36)(u - 1). (I'm more used to using u for substitutions.)

With the substitutions, we have
[tex]SA = \pi/18 \int_{u = 1}^{37} [2 - (1/36)(u - 1)]u^{1/2}du[/tex]
[tex]= \pi/18 \int_{u = 1}^{37} [2u^{1/2} - (1/36)u^{3/2} + (1/36)u^{1/2}]du[/tex]

By omitting dx, you neglected a factor of 1/36 with du (since dx = (1/36)du).
I got x= (1/36)(u-1) but then I tried to factor out the (1/36) to make integration simpler, and that seems to be where I messed up. I'll just continue without trying to factor it out.

Thanks! :)
 

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