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

- anonymous

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

- schrodinger

See more answers at brainly.com

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

what is the question?

- anonymous

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

- anonymous

And what was your answer?

Looking for something else?

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

## More answers

- anonymous

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

- anonymous

-5/6

- anonymous

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

- anonymous

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

- anonymous

Oh. Yes I also got -5/6

- anonymous

Thanks!

- anonymous

No it is not -5/6

- anonymous

opps..ya it is, sorry

- anonymous

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

- anonymous

:)

- 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?

- anonymous

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

- anonymous

Haha, Ok here goes:

- anonymous

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

- anonymous

Convert to polar.

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

is it the x=rcostheta and y=rsintheta?

- anonymous

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

- 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.

- anonymous

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

- anonymous

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

- 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.

- 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.

- anonymous

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

- 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.

- 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!

Looking for something else?

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