Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

change to polar and then integrate

Mathematics
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.

Join Brainly to access

this expert answer

SIGN UP FOR FREE
\[\int\limits_{0}^{2}\int\limits_{0}^{\sqrt{4-y ^{2}}} x ^{2}+y^{2} dxdy\]
i know it becomes \[r^{2} r dr dTheta\]
yes, which simplifies to\[r^3drd\theta\]where are you stuck?

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

the sqrt term becomes x^2+y^2= 4 right
x^2+y^2=r^2 what is the radius of the circle in question?
so i should get r^4/4 evaluated from 0 to 2 then i should get 4 dtheta
radius should be 2
yes
and what are the bounds on theta?
0 to 2pi
remember that sqrt(4-y^2) is only the *top* of the circle, the bottom would require a negative sign, so the bounds on theta are only...?
0 to pi
yes
so would i ge 4 pi
yes
ur a life saver
no problem, happy to help

Not the answer you are looking for?

Search for more explanations.

Ask your own question