C4 integration help Watch

hjkl1996
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#1
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The region under the graph of y=lnx between x-1 and x=3 is rotated through 2pi radians about the x-axis to from a solid revolution.
b) by using substitution x=e^u, find the exact value of the volume
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james22
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What have you done so far?
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the bear
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you need to integrate y2dx and times by pi
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hjkl1996
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(Original post by james22)
What have you done so far?
I've got that x^2=e^2u and dx=ue^u du so it's pi ∫ ln (e^2u) ue^u du
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james22
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(Original post by hjkl1996)
I've got that x^2=e^2u and dx=ue^u du so it's pi ∫ ln (e^2u) ue^u du
You have to integrate y^2 not x^2
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interstitial
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Do you know the formula for revolution around the x axis?

Integrate y^2 and multiply by 2Π

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hjkl1996
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(Original post by james22)
You have to integrate y^2 not x^2
so it's ln(e^u)^2 ue^u du? How do you integrate that?
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interstitial
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(Original post by hjkl1996)
so it's ln(e^u)^2 ue^u du? How do you integrate that?
Wouldn't it just be \displaystyle\int_1^3 ln^2(x) \: \mathrm d x

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hjkl1996
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but you have to write x and dx in terms of u
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interstitial
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(Original post by hjkl1996)
but you have to write x and dx in terms of u
Then x^2=(e^u)^2=(e^2)^{u^2}

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james22
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Not quite.
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hjkl1996
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(Original post by james22)
Not quite.
so then how do you integrate the whole thing? Btw the answer is pi[3ln(3)^2 -6ln3+4]
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james22
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(Original post by hjkl1996)
so then how do you integrate the whole thing? Btw the answer is pi[3ln(3)^2 -6ln3+4]
You made an error in calculating dx, you have x=e^u so dx/du=e^u so dx=e^u du
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hjkl1996
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(Original post by james22)
You made an error in calculating dx, you have x=e^u so dx/du=e^u so dx=e^u du
Oh yeah, thanks! So now would you integrate by parts?
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james22
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(Original post by hjkl1996)
Oh yeah, thanks! So now would you integrate by parts?
Yes. In the OP you put b) before the question, I am guessing that there was a part a) which may help. What was the full question?
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