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Andresfon12
 2 years ago
evaluate the triple integral
Andresfon12
 2 years ago
evaluate the triple integral

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Andresfon12
 2 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{0}^{5} \int\limits_{0}^{1} \int\limits_{0}^{\sqrt {1z^2}} ze^y dx dz dy\]

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.1what's the problem you're having?

Andresfon12
 2 years ago
Best ResponseYou've already chosen the best response.0inter sqrt 0_1z^2 and finding dx

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.1\[\int dx=x\]so\[\int_0^5\int_0^1\int_0^{\sqrt{1z^2}}ze^ydxdzdy=\int_0^5\int_0^1\left.ze^yx\right]_0^{\sqrt{1z^2}}dzdy\]\[=\int_0^5\int_0^1ze^y\sqrt{1z^2}dzdy\]and keep working from the inside out...

Andresfon12
 2 years ago
Best ResponseYou've already chosen the best response.0that mean \[\int\limits_{0}^{5} \int\limits_{0}^{1} z^2/2e^y *2/3 (1z^2) z ? dy\]

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.1all the parts not dependent on z can be taken out of the inner integral and treated as constants\[\int_0^5\int_0^1ze^y\sqrt{1z^2}dzdy=\int_0^5e^y\left[\int_0^1z\sqrt{1z^2}dz\right]dy\]what is this inner integral?

Andresfon12
 2 years ago
Best ResponseYou've already chosen the best response.0i guess i can substitution let u= 1z^2 du=2z dz => du/2 =z dz
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