- #1
trancefishy
- 75
- 0
ok, so I've got this triple integral: [tex] \int_{0}^{2} \int_{0}^{y^3} \int_{0}^{y^2} f(x,y,z) dz\, dx\, dy\ [/tex]
what i want to do is get the other five integrals that are equivalen. I've got correctly 3 of them, but, for the life of me, cannot get dy dz dx and dy dx dz to work out.
i've bumbled around with [tex] \int_{0}^{8} \int_{0}^{x^{2/3}} \int_{x^{1/3}}^{2} f(x,y,z) dy\, dz\, dx\ [/tex]
and
[tex] \int_{0}^{8} \int_{0}^{4} \int_{x/z}^{2} f(x,y,z) dy\, dz\, dx\ [/tex]
among many others, and can't seem to get it pegged down.
if i integrate to check (just having a constant in the beginning) the correct answer seems to be 32/3. anyways, ANY clues to this, would help me a ton. I'm positive I'm missing something blatant or just have the region thought out incorrectly.
what i want to do is get the other five integrals that are equivalen. I've got correctly 3 of them, but, for the life of me, cannot get dy dz dx and dy dx dz to work out.
i've bumbled around with [tex] \int_{0}^{8} \int_{0}^{x^{2/3}} \int_{x^{1/3}}^{2} f(x,y,z) dy\, dz\, dx\ [/tex]
and
[tex] \int_{0}^{8} \int_{0}^{4} \int_{x/z}^{2} f(x,y,z) dy\, dz\, dx\ [/tex]
among many others, and can't seem to get it pegged down.
if i integrate to check (just having a constant in the beginning) the correct answer seems to be 32/3. anyways, ANY clues to this, would help me a ton. I'm positive I'm missing something blatant or just have the region thought out incorrectly.