Here's the question you clicked on:
Andresfon12
evaluate the triple integral
\[\int\limits_{0}^{5} \int\limits_{0}^{1} \int\limits_{0}^{\sqrt {1-z^2}} ze^y dx dz dy\]
what's the problem you're having?
inter sqrt 0_1-z^2 and finding dx
\[\int dx=x\]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...
that mean \[\int\limits_{0}^{5} \int\limits_{0}^{1} z^2/2e^y *2/3 (1-z^2) z ? dy\]
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?
i guess i can substitution let u= 1-z^2 du=2z dz => du/2 =z dz