Triple Integrals: Solving \int\int\int^{}_{B} ye^(-xy) dV

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The discussion revolves around solving the triple integral ∫∫∫_B ye^(-xy) dV, where B is defined by the bounds 0 ≤ x ≤ 4, 0 ≤ y ≤ 1, and 0 ≤ z ≤ 5. Initial attempts at integration lead to expressions involving logarithms, which create complications when evaluating at the lower limit of zero. Participants suggest switching the order of integration to simplify the process, which results in a more manageable expression. Despite this, errors in calculations are noted, particularly when evaluating at the bounds, leading to incorrect results. The conversation emphasizes careful execution of integration steps to avoid mistakes.
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Homework Statement


\int\int\int^{}_{B} ye^(-xy) dV where B is the box determined by 0 \leq x \leq 4, 0 \leq y \leq 1, 0 \leq z \leq 5.

Homework Equations


The Attempt at a Solution


\int^{4}_{0}\int^{1}_{0}\int^{5}_{0} ye^(-xy) dzdydx
Integrating the first time I get
zye-xy
Plugging in 5 and 0 I get
5ye-xy
Integrating the above with respect to y. I use u = 5y and dv = e-xy which gives me du = 5du and v = \frac{-e^(-xy)}{x}
Which leaves me with the following equation.
-5y*\frac{e^(-xy)}{x} - \int e^(-xy)5du
After integration I get
-5y*\frac{e^(-xy)}{x} + \frac{5e^(-xy)}{x}
Plugging in 1 and 0 into the above I get
-5\frac{e^(-x)}{x} + 5\frac{e^(-x)}{x} - 5\frac{e^0}{x}
Which just leaves me with since the first two cancel each other out.
-5\frac{e^0}{x}
Integrating the above I get
-5log(x) which is where my problem lies, I can't get the log of 0.
 
Last edited:
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Let's fix up the latex a little. For exponents, use e^{-xy} instead of e^(-xy).

Homework Statement



\int\int\int^{}_{B} ye^{-xy} dV where B is the box determined by 0 \leq x \leq 4, 0 \leq y \leq 1, 0 \leq z \leq 5.


Homework Equations




The Attempt at a Solution



\int_0^4 \int_0^ 1 \int_0^5 ye^{-xy} dzdydx

= \int_0^4 \int_0^1 \Bigr|_0^5 zye^{-xy} dzdydx

= \int_0^4 \int_0^1 5ye^{-xy} dydx


At this point, I think it would be simpler to integrate with respect to x.
 
Didn't even think about that . . .but for some reason it doesn't seem to be working.
I switched the integration around and got the following.
\int_0^1 \int_0^4 5ye^{-xy} dxdy
After the first integration it comes out really neatly as
-5e-xy
and after plugging in 0 and 4 I get
-5e-4y
Integrating the above I get
5/4e-4y
but after plugging in 0 and 1 I get the wrong answer, not sure what I am doing wrong.
 
shards5 said:
Didn't even think about that . . .but for some reason it doesn't seem to be working.
I switched the integration around and got the following.
\int_0^1 \int_0^4 5ye^{-xy} dxdy
After the first integration it comes out really neatly as
-5e-xy
and after plugging in 0 and 4 I get
-5e-4y

5 e^{-(0)(y)} isn't zero.
 
I guess I shouldn't have rushed through the calculations as I did, thanks a lot for pointing out my mistake.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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