I'm only 16 and I am doing some AP calculus problems... Anyone good or somewhat okay at it?

- Mitchi35

I'm only 16 and I am doing some AP calculus problems... Anyone good or somewhat okay at it?

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- chestercat

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- sleepyhead314

We got plenty of people that can help you here :)
I got a 5 on both the AB and BC calc tests ^_^

- Mitchi35

Oh wow! Okay I'm sold. Let me just put the problem up real quick.

- Mitchi35

Give the volume of the solid generated by revolving the region bounded by the graph of y = ln(x), the x-axis, the lines x = 1 and x = e, about the y-axis.

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## More answers

- anonymous

What is the derivatives of:
\[x^{-3}\]
and
\[ e^{2x}\]
with respect to x?

- sleepyhead314

haha @PlasmaFuzer I would suggest that you ask your own questions in the "Ask a question..." box to the left ^_^

- anonymous

Ooo I like the question even better @Mitchi35

- anonymous

@sleepyhead314 I thought he was asking for us to quiz him... I have many more where that came from, but I guess I just misunderstood what he was looking for

- Mitchi35

Lol What are yo talking about? @PlasmaFuzer

- Mitchi35

Oh Okay. :)

- sleepyhead314

I see, it's fine xD
but feel free to jump in to help anytime @PlasmaFuzer :)

- anonymous

I didn't see your second post until I had already posted mine, and then I confused you for the the other person in here and not the poster XD sorry.

- sleepyhead314

now, I'm guessing you Are allowed a graphing calculator for this problem? @Mitchi35

- anonymous

I would @sleepyhead314 but I am afraid its been quite a while for me on these types of problems so I think I might just cheat step my way to the answer which isn't instructive if you don't understand why things work the way they do

- anonymous

I.e. just integrate and rotate

- Mitchi35

And I'm a she by the way.@ PlasmaFuzer. And Yes. But only I don't have one. @sleepyhead314

- sleepyhead314

|dw:1444067181310:dw|

- anonymous

Sorry again XD

- Mitchi35

Oh wait! They have multiple choice. I"ll put them up.

- anonymous

I don't want to step on any toes @sleepyhead314 but I am also curious to see if I remembered correctly can I give it a shot and if I am wrong then you can correct?

- sleepyhead314

of course @PlasmaFuzer feel free to take the wheel ^_^

- anonymous

I am mostly just debating between two approaches, my gut tells me you calculate the integral of the function (i.e. area under curve), then you rotate that 2pi radians about the x-axis....

- anonymous

Then again I have to consider that the infinitesimal areal elements are circles of radius ln(x) and so my integrand should be ln(x)^2 (i.e. imitate the circle formula)

- anonymous

times pi

- anonymous

Considering further I think it should be the first, because the integral derivation of the area of a circle involves the circumference as the integrand.... therefore my first approach I believe to be correct.

- jigglypuff314

try the disk method,
volume = pi times the integral of the function from bound to bound squared

- jigglypuff314

in this case
your function is ln(x)
and your bounds are from 1 to e

- anonymous

Ahh yes ty jiggly Its the square of the integral not the integrand squared that makes more sense.

- anonymous

You can take this one Im obviously rusty

- Mitchi35

Yeah. It's fine guys. I'll just ask the teacher. Thank you very much though anyways. :)

- anonymous

Oh @Mitchi35 I can help if they left.... I realized my mistake, and if they aren't there to help finish the problem I am more than able.

- anonymous

I just was gonna give them first crack since I kinda screwed up.... Anyways think of the problem as assembling a bunch of disks whose area is given by pi*r^2 where the radius in this instance will be an really thing slice of the area under the curve.

- anonymous

Ok @Mitchi35 I was right before I just pulled up a quick review site and it isnt the integral squard as jigglypuff suggested but it is the INTEGRAND squared prior to integration. Give me a moment and I will type up a solution.

- jigglypuff314

sorry if our explanations are a little confusing @Mitchi35
I'll attach the answer I got for your question on this comment for you
hopefully that might help a bit more :)

##### 1 Attachment

- anonymous

Oh yeah @jigglypuff314 if thats what you meant I totally agree.... I was torn between my first two approaches and thought what you said was implying a third.... but really it was just confirming the second approach.

- jigglypuff314

I was simply using the approach that my BC calc teacher taught me :P

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