Triple Integral

[renegade05]

Question:

\(\displaystyle \int_0^4 \! \int_x^4 \! \int_0^y \! \frac{6}{1+48z-z^3} \, \mathrm{d} z\mathrm{d} y\mathrm{d} x\)

Integrate by hand.

Ugh, this one is giving me a headache. I am trying to draw the region that this is bounded to maybe switch the order of integration, but I'm having a very hard time!

Any tips on how to approach this one? I even though about switching over to cylindrical or spherical coordinates, but i don't think that will do much good.

Thanks!

 

[renegade05]

Never mind, i figured it out.

\(\displaystyle \int_0^4 \! \int_x^4 \! \int_0^y \! \frac{6}{1+48z-z^3} \, \mathrm{d} z\mathrm{d} y\mathrm{d} x

~=

\int_0^4 \! \int_z^4 \! \int_0^y \! \frac{6}{1+48z-z^3} \, \mathrm{d} x\mathrm{d} y\mathrm{d} z ~ =ln(129)\)

I would delete this thread, but I don't know how.