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

[shards5]

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.

 

[Stephen Tashi]

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.

 

[shards5]

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.

 

[Stephen Tashi]

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.

 

[shards5]

I guess I shouldn't have rushed through the calculations as I did, thanks a lot for pointing out my mistake.