Question

Anyone willing to verify my answer to a double integral? Thanks

Answer 1

what is the question?

Answer 2

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

Answer 3

And what was your answer?

Answer 4

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

Answer 5

-5/6

Answer 6

You want us to verify your answer you should give your answer ;p

Answer 7

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

Answer 8

Oh. Yes I also got -5/6

Answer 9

Thanks!

Answer 10

No it is not -5/6

Answer 11

opps..ya it is, sorry

Answer 12

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

Answer 13

:)

Answer 14

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?

Answer 15

If you ask, we may answer. If you don't ask we certainly will not help ;p

Answer 16

Haha, Ok here goes:

Answer 17

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

Answer 18

Convert to polar.

Answer 19

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

Answer 20

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

Answer 21

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

Answer 22

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

Answer 23

is it the x=rcostheta and y=rsintheta?

Answer 24

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

Answer 25

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.

Answer 26

So your radius goes from 0 to 1, and your angle goes from 0 to Pi.

Answer 27

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

Answer 28

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.

Answer 29

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.

Answer 30

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

Answer 31

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.

Answer 32

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!