Double Integral Problem

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November 29th 2009, 12:53 PM
BCHurricane89

Double Integral Problem

Hello, I am having difficulty setting up this double integral.

$f(s,t)=e^slnt$ over the region in the first quadrant of the st plane that lies above the curve $s=lnt$ from $t=1$ to $t=2$

so far I have converted the $s=lnt$ into the format: $t=e^s$ Then I graphed it, and for my bound on my integral, I got: $\int_1^2\int_o^{e^s}{e^slnt}$ $dtds$

Nevermind....after trying numerous different things, I think I have finally found the right answer. The x bounds were really getting me. I knew the y bounds had to be 1 and 2, since that was given, but for some reason, the x bounds were giving me trouble.

$\int_1^2\int_o^{lny}{e^slnt}$ $dtds$
November 29th 2009, 11:21 PM
simplependulum

Quote:

Originally Posted by BCHurricane89

$\int_1^2\int_o^{e^s}{e^slnt}$ $dtds$

It is wrong ( just my view )

I think the correct boundary is

$1 \leq t \leq 2 , 0 \leq s \leq <br /> ln(t)$

so what we have to calculate is

$\int_1^2\int_0^{\ln(t)} e^s \ln(t)~dsdt$

integrate w.r.t. $s$ first .

$= \int_1^2 \ln(t) [ e^s ]_0^{\ln(t)} ~dt$

$= \int_1^2 (t-1)\ln(t)~dt = [(t-1)^2\ln(t)/2 - t^2/4 + t - \ln(t)/2 ]_1^2$

$= 2 - 1 + 1/4 -1 = 1/4$
November 29th 2009, 11:35 PM
simplependulum

- Wolfram|Alpha

It tells us the answer but no steps are shown ( pity ) .

I think it is easy to input ... so why not you check your exercises online (Rofl)

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