anonymous 5 years ago Anyone willing to verify my answer to a double integral? Thanks

1. anonymous

what is the question?

2. anonymous

$\int\limits_{-3}^{2}\int\limits_{0}^{1}y^{2} x$dy dx

3. anonymous

4. anonymous

Should I write it? Might it influence others' solution?

5. anonymous

-5/6

6. anonymous

7. anonymous

Lol, I meant see what answer you get and then I'll check if it's the same.

8. anonymous

Oh. Yes I also got -5/6

9. anonymous

Thanks!

10. anonymous

No it is not -5/6

11. anonymous

opps..ya it is, sorry

12. anonymous

Yeah I got something different at first, but I thought it was dxdy, not dydx

13. anonymous

:)

14. anonymous

lol! Thanks! Just making sure I know what I'm doing... I've got another similar question which I haven't answered yet as I need help answering it. Want to help out?

15. anonymous

16. anonymous

Haha, Ok here goes:

17. anonymous

$\int\limits_{-1}^{1}\int\limits_{0}^{\sqrt{1-x ^{2}}}(x ^{2}+y ^{2})^{3/2}$dydx

18. anonymous

Convert to polar.

19. anonymous

$\int\limits_{0}^{\pi}\int\limits_{0}^{1}r^3 r\ drd \Theta$

20. anonymous

Does that make sense? I'm getting Pi/5 for the answer.

21. anonymous

Thanks dude, sorry about the late reply, my internet went and has finally JUST come back!

22. anonymous

Can you help me a bit step by step... Thanks. How did you convert to polar?

23. anonymous

is it the x=rcostheta and y=rsintheta?

24. anonymous

$r^2 = x^2 + y^2 \rightarrow r = (x^2 + y^2)^{1/2} \rightarrow (x^2 + y^2)^{3/2} = r^3$ $dydx = r\ drd \Theta$

25. anonymous

Then you have to convert your limits. Since x goes from -1 to 1 and y goes from 0 to sqrt(1+x^2) you can see that you're integrating over the top half of a circle with radius 1.

26. anonymous

27. anonymous

Can you explain how you realised that it is the top half of a circle?

28. anonymous

Because the y values go from 0 to sqrt(1+x^2) Those are all positive y values inside the circle x^2 + y^2 = 1 If it had said -sqrt(1+x^2) then it would have been the bottom half of the circle.

29. anonymous

If you graph the line y = sqrt(1+x^2) you'll see it forms the top half of a circle. Integrating from y = 0 to y=sqrt(1+x^2) will be integrating over that same half-circle.

30. anonymous

I understand the 0 to 1 limit of y. How did the 0 to Pi limit come about? Thanks

31. anonymous

You have to look at both pieces together. If y goes from 0 to the top of the half circle and x goes from -1 to 1 then the area you're covering is the top half of the circle. That means your r will go from 0 to 1 and your Theta will go from 0 to pi. cos(0) = 1 = x, cos(pi) = -1 = x.

32. anonymous

Oh! Sorry about that! Theta goes from 0 to Pi, I was thinking about x... Understand that part now! Thanks a lot for your help!