anonymous
  • anonymous
evaluate the triple integral
Mathematics
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

anonymous
  • anonymous
\[\int\limits_{0}^{5} \int\limits_{0}^{1} \int\limits_{0}^{\sqrt {1-z^2}} ze^y dx dz dy\]
TuringTest
  • TuringTest
what's the problem you're having?
anonymous
  • anonymous
inter sqrt 0_1-z^2 and finding dx

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

TuringTest
  • TuringTest
\[\int dx=x\]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...
anonymous
  • anonymous
that mean \[\int\limits_{0}^{5} \int\limits_{0}^{1} z^2/2e^y *2/3 (1-z^2) z ? dy\]
TuringTest
  • TuringTest
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?
anonymous
  • anonymous
i guess i can substitution let u= 1-z^2 du=2z dz => du/2 =z dz
anonymous
  • anonymous
use*

Looking for something else?

Not the answer you are looking for? Search for more explanations.