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**hristo****Member**- Registered: 2005-12-26
- Posts: 8

hi,

i have a problem with a volume question. it's the following:

Let R be the shaded region in quadrant 1 bounded by the graphs of y = ln x and y = 1.

find the volume of the solid generated by rotating R about the y-axis.

book-answer: pi/2 (e^2-1)

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**John E. Franklin****Member**- Registered: 2005-08-29
- Posts: 3,588

y = ln x

Draw a graph and note are going to integrate from

y = 0 to y = 1.

because of pi r^2 for area. (disk method, might be called)

From an integral table:

Great answer, but I wish I knew how to

do the integral without the table lookup.

*Last edited by John E. Franklin (2006-01-13 16:33:02)*

**igloo** **myrtilles** **fourmis**

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**hristo****Member**- Registered: 2005-12-26
- Posts: 8

but what happens with the y = 1? i need it to find the limits. it can't be the upper limit because the graphs don't intersect there. they intersect when x = e in this equation: y = ln x, so the upper limit should be e. correct me if i am wrong and please tell me why i am wrong.

*Last edited by hristo (2006-01-13 16:43:47)*

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**John E. Franklin****Member**- Registered: 2005-08-29
- Posts: 3,588

Click on graph to make bigger.

**igloo** **myrtilles** **fourmis**

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**John E. Franklin****Member**- Registered: 2005-08-29
- Posts: 3,588

I am integrating along the y-axis, not the x-axis.

**igloo** **myrtilles** **fourmis**

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**John E. Franklin****Member**- Registered: 2005-08-29
- Posts: 3,588

For each microscopic y position, imagine a thin disc

that goes around the y-axis, so the discs are all

stacked up like coins if y is up and x is to the right.

**igloo** **myrtilles** **fourmis**

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**hristo****Member**- Registered: 2005-12-26
- Posts: 8

ohh, ok. thanks. i feel like a retard now. hehe

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**John E. Franklin****Member**- Registered: 2005-08-29
- Posts: 3,588

No problem. This is all fairly new to me, but hopefully someone can

show how to integrate without the table.

**igloo** **myrtilles** **fourmis**

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

That's just integration by recognition. We know that if you differentiated

, you'd get .So, if we want the answer to be half of that, we need to half the thing we're differentiating.

Using this backwards shows that

.Why did the vector cross the road?

It wanted to be normal.

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**irspow****Member**- Registered: 2005-11-24
- Posts: 457

V = ∫2πx(f(x)-g(x))dx

2π∫x - xlnx dx = π[x²]0,e - 2π[x²lnx / 2]1,e + 2π[x²/4]1,e

Note that I had to use integration by parts for g(x).

V = πe² - πe² + 2π(e²/4 - 1/4) = π/2 (e²-1) ≈ 10.0359u²

Also note the upper and lower limits for the integrations above. We had to exclude the values of x lower than 1 for the logarithmic part because that would have been negative and subtracted from the shaded region above the x axis. I hope that this clears up some of the questions above.

John, note that even though you attempted to use the disk method instead of the cylindrical shell method that the volume you tried to compute was for the region below the lnx function. If you were to use the disk method the integral above should be subtracted from the integral of y=1 function.

*Last edited by irspow (2006-01-14 03:32:36)*

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