A community for students. Sign up today!
Here's the question you clicked on:
 0 viewing
 2 years ago
Using geometry (the interpretation of a double integral as a volume), evaluate the double integral
∫∫D sqrt(16−x^2−y^2) dA over the circular disk D: x^2+y^2≤16.
 2 years ago
Using geometry (the interpretation of a double integral as a volume), evaluate the double integral ∫∫D sqrt(16−x^2−y^2) dA over the circular disk D: x^2+y^2≤16.

This Question is Closed

nubeer
 2 years ago
Best ResponseYou've already chosen the best response.0hmm you can convert this in polar coordinates right?

nubeer
 2 years ago
Best ResponseYou've already chosen the best response.0x=rcostheta y = r sin theata if i am not wrong.. limit of r would be 04 and theta limit 02pi

malevolence19
 2 years ago
Best ResponseYou've already chosen the best response.1Yeah, so remember: \[dA= \mathcal{J}(r,\phi)dr d \phi=r dr d \phi\] \[x=r \cos \phi; y=r \sin \phi\] \[D: \left\{ r^2 \le 16 \right\} \implies D: \left\{ r \le 4 \right\}\] Full circle implies \[0 \le \phi \le 2 \pi\] So your integral would be: \[\int\limits_0^{2 \pi} \int\limits_0^4 (16r^2)r dr d \phi\] As: \[r^2=x^2+y^2\]

zhiyuan3yu5
 2 years ago
Best ResponseYou've already chosen the best response.0what do i do with the square root? also the problem says to use geometry. but i'm not sure what the height would be

cnknd
 2 years ago
Best ResponseYou've already chosen the best response.0ok let z = ur integrand = sqrt(16x^2y^2) what's the geometric relationship between x,y,and z?

cnknd
 2 years ago
Best ResponseYou've already chosen the best response.0k one more hint: i said z = sqrt(16x^2y^2). let me rewrite this as: x^2+y^2+z^2 = 16 (with the understanding that z can't be negative). does this equation ring a bell?
Ask your own question
Ask a QuestionFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.