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**mikau****Member**- Registered: 2005-08-22
- Posts: 1,504

Let R be the region under the graph of y = tan x and the x axis from x = 0 to x = pi/2 (90 degree's)

1. Find the area of the R.

2. Find the volume of the solid formed when R is rotated about the y axis.

1. This is easy. Integrate tan x dx you get - ln cos x or ln 1/cos x, from x = 0 to x = pi/2.

limit of ln 1/cos b - ln 1/cos 0 as b approaches pi/2 from the left. ln 1/cos 0 = ln 1 = 0 so we can ignore that part. As b approcaches 90 from the left, cos b approaches zero (from the positive side), so ln 1/cos 0+ = ln ∞. So its infinite area.

2. Now this is where I'm stuck. Finding the volume of the solid when R is rotated about the y axis. Using the shell method, the radius will be x, so the circumfrence of each ring is 2pi x, the height of each ring is tan x, and the thickness is dx.

2pi ∫ x tan x dx from x = 0 to x = pi/2

I cannot for the life of me, integrate this. Using integration by parts doesn't seem to help. If we choose to let u = x, and dV = tan x, we have to to integrate ln|cos x| dx and I don't think this can be integrated since the derivative of cos x is not present.

If we instead choose to let dV = x dx, and u = tan x, that seems to make it worse, if you try integration by parts more then once, x becomes 1/2 x^2 then 1/6 x^3 then 1/24 x^4 and so on. While differentiating tan x just produces more and more trigonometrc functions and nothing is ever eliminated.

However, it seems obvious that if the area of tan x from 0 to pi/2 is infinity, then the volume is definily infinity.

2 pi ∫ x tan x dx >= ∫ tan x dx for all values of x between 0 and pi/2. Since ∫ tan x dx from 0 to pi/2 diverges, the other should also diverge. But I learned this comparison test for series and don't know if its legal to apply it to integrals.

Anyways, clearly its infinity but I do believe the book wants me to integrate and evaluate the limit as b approaches pi/2. But I can't seem to integrate ∫ x tan x dx :-(

Advice?

A logarithm is just a misspelled algorithm.

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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,905

Mathematica tells me that this inegral does not coverage. It's "oo".

IPBLE: Increasing Performance By Lowering Expectations.

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**mikau****Member**- Registered: 2005-08-22
- Posts: 1,504

don't you mean does not *converge*? Meaning it diverges? That seems obvious to me, but I'm just wondering if I should simply say it converges by comparison, or integrate and evaluate, in which case, I'm stuck.

A logarithm is just a misspelled algorithm.

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**mikau****Member**- Registered: 2005-08-22
- Posts: 1,504

nevermind, I think I got it. The trick is to let tan x = sin x / cos x and let x sin x = dV and 1/cos x = u.

(edit) Oops. Thats not it. I did it wrong.

So I still need help.

*Last edited by mikau (2006-03-05 08:57:31)*

A logarithm is just a misspelled algorithm.

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**mikau****Member**- Registered: 2005-08-22
- Posts: 1,504

sooo...any suggestions? Should I say it diverges by comparison or keep trying to integrate?

A logarithm is just a misspelled algorithm.

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**George,Y****Member**- Registered: 2006-03-12
- Posts: 1,316

Integrate ((π/2)[sup]2[/sup]-[arctan(y)][sup]2[/sup]) dy from 0 to infinity, i guess.

*Last edited by George,Y (2006-03-20 01:25:50)*

**X'(y-Xβ)=0**

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