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October 18, 2025, 01:59:07 pm
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Author Topic: dejan91's questions  (Read 17035 times)  Share 

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dejan91

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Re: dejan91's questions
« Reply #60 on: June 15, 2009, 11:33:36 pm »
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Yeah knew integration by parts was the other way. Just thought there was a different method.

, and area of rectangle is . Is that right?

Is it worth learning integration by parts by the way?
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TrueTears

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Re: dejan91's questions
« Reply #61 on: June 15, 2009, 11:37:20 pm »
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Yeah, it's not very hard to learn, a quick wikipedia can help :) [Good knowledge to have especially when doing integration, even though it's not in spesh course.]

Area of the rectangle is

You integral for the inverse is fine, all you gotta do now is (Result of the integral of the inverse) which yields the area required.
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dejan91

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Re: dejan91's questions
« Reply #62 on: June 15, 2009, 11:42:13 pm »
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Oh, crap. No wonder I got it wrong. This is what I was getting confused about: what values do you take for the side lengths of the rectangle? The maximum for both the x and y limits (in this case 2 and log_e(4)? What if the lower limit was x=1 or something?

Hmmm, might have a look into that integration by parts.
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TrueTears

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Re: dejan91's questions
« Reply #63 on: June 15, 2009, 11:53:47 pm »
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Doesn't matter what the lower limit is, you are right, it is just the upper limit and its respective y co ordinate.

To understand it graphically, when you integrate with respect to y, you are working out the area bounded by the graph and the y axis. Hence you use the rectangle minus that area to get the area required.

Here's are good links to learn integration by parts: (This is how I learnt it)

1. http://www.youtube.com/watch?v=WRKMunWZqU4

2. http://en.wikipedia.org/wiki/Integration_by_parts
« Last Edit: June 15, 2009, 11:58:33 pm by TrueTears »
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dejan91

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Re: dejan91's questions
« Reply #64 on: June 17, 2009, 06:48:21 pm »
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The region between the curves , and the lines and is rotated about the x-axis. Find the volume of the solid formed.

I've got the volume as:

Not sure what to do with the though. Help please?


EDIT: karma 22, posts 222. LOL.
« Last Edit: June 17, 2009, 06:51:07 pm by dejan91 »
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Re: dejan91's questions
« Reply #65 on: June 17, 2009, 06:50:29 pm »
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sub in 2x for x.

Then solve for
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dejan91

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Re: dejan91's questions
« Reply #66 on: June 17, 2009, 06:56:18 pm »
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Do you mean to do ?
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TrueTears

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Re: dejan91's questions
« Reply #67 on: June 17, 2009, 07:00:51 pm »
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I mean this





Sub that back in :)
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dejan91

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Re: dejan91's questions
« Reply #68 on: June 17, 2009, 07:04:17 pm »
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Ahhh of course. Forgot you could do that! Thank you!
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dejan91

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Re: dejan91's questions
« Reply #69 on: July 03, 2009, 12:14:09 am »
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A conical flask with a raduis of 5cm and a height of 12cm is being filled with powder at a constant rate of . Find the rate, in terms of height, at which the height of the powder is changing.
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Re: dejan91's questions
« Reply #70 on: July 03, 2009, 12:34:19 am »
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dv/dt = pi


volume of the cone is:


but oh wait, its being filled from the pointy end, so

, where is the height of the powder measured from the bottom of the cone. is the height of the flask.  <- draw a diagram

find the ratio for to ,     (draw a diagram)
hence get an equation for h in terms of r
put this into the above equation,
find dh/dv,
chain rule it.
« Last Edit: July 03, 2009, 12:43:29 am by zzdfa »
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dejan91

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Re: dejan91's questions
« Reply #71 on: July 03, 2009, 09:27:28 am »
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Thanks :)
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dejan91

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Re: dejan91's questions
« Reply #72 on: October 31, 2009, 09:34:40 pm »
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The curve formed by the graph of is rotated around the y-axis from to where is a real constant, , to form the shape of a funnel. and are measured in cm. Find,in exact terms, the volume of the funnel for .
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TonyHem

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Re: dejan91's questions
« Reply #73 on: October 31, 2009, 09:43:22 pm »
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Does the answer come out nicely? I wrote a few things up and it looks like it's gonna come out with heaps of terms :/
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dejan91

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Re: dejan91's questions
« Reply #74 on: October 31, 2009, 09:49:10 pm »
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Well, answer is: . Is that what you got?
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