Question

evaluate the triple integral

Answer 1

\[\int\limits_{0}^{5} \int\limits_{0}^{1} \int\limits_{0}^{\sqrt {1-z^2}} ze^y dx dz dy\]

Answer 2

what's the problem you're having?

Answer 3

inter sqrt 0_1-z^2 and finding dx

Answer 4

\[\int dx=x\]so\[\int_0^5\int_0^1\int_0^{\sqrt{1-z^2}}ze^ydxdzdy=\int_0^5\int_0^1\left.ze^yx\right]_0^{\sqrt{1-z^2}}dzdy\]\[=\int_0^5\int_0^1ze^y\sqrt{1-z^2}dzdy\]and keep working from the inside out...

Answer 5

that mean \[\int\limits_{0}^{5} \int\limits_{0}^{1} z^2/2e^y *2/3 (1-z^2) z ? dy\]

Answer 6

all the parts not dependent on z can be taken out of the inner integral and treated as constants\[\int_0^5\int_0^1ze^y\sqrt{1-z^2}dzdy=\int_0^5e^y\left[\int_0^1z\sqrt{1-z^2}dz\right]dy\]what is this inner integral?

Answer 7

i guess i can substitution let u= 1-z^2
du=2z dz => du/2 =z dz

Answer 8

use*