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C4 integration help

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
Reply 1
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james22
16
What have you done so far?
Reply 2
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the bear
20
you need to integrate y2dx and times by pi
Reply 3
Avatar for hjkl1996
hjkl1996
OP
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
Reply 4
Avatar for james22
james22
16
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
Do you know the formula for revolution around the x axis?

Integrate y^2 and multiply by 2Π

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Reply 6
Avatar for hjkl1996
hjkl1996
OP
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?
Original post by hjkl1996
so it's ln(e^u)^2 ue^u du? How do you integrate that?


Wouldn't it just be ∫13ln2(x) dx\displaystyle\int_1^3 ln^2(x) \: \mathrm d x

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Reply 8
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hjkl1996
OP
but you have to write x and dx in terms of u
Original post by hjkl1996
but you have to write x and dx in terms of u


Then x2=(eu)2=(e2)u2x^2=(e^u)^2=(e^2)^{u^2}

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Reply 10
Avatar for james22
james22
16
Original post by majmuh24
Then x2=(eu)2=(e2)u2x^2=(e^u)^2=(e^2)^{u^2}

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Not quite.
Reply 11
Avatar for hjkl1996
hjkl1996
OP
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]
Reply 12
Avatar for james22
james22
16
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
Reply 13
Avatar for hjkl1996
hjkl1996
OP
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?
Reply 14
Avatar for james22
james22
16
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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