double integeration xy dxdy over poasitive quadrant of circle
???

Question

double integeration xy dxdy over poasitive quadrant of circle
???

turingtest · Answer

the radius of the circle would help

turingtest · Answer

you are given the radius, right?

turingtest · Answer

@chakshu if you don't provide all the info in the question we can't help you. You must be either given the formula for the circle or its radius.

anonymous · Answer

Hint #1: What @TuringTest said.
Hint #2: The double-integral is the area under the curve.

anonymous · Answer

@TuringTest I believe @chakshu wants the double-integral of a circle over one quadrant - 0 to pi/2.

turingtest · Answer

clearly, yeah (I was gonna make him tell me those bounds on theta, but oh well) but we certainly need the radius.

anonymous · Answer

the question is based on double integeration of xy with respct to dxdy for positive quadrant of circle x^2+y^2=a^2...hope u understand??

turingtest · Answer

ok so the radius is "a" then, which we can leave unknown

do you know how to convert rectangular to polar coordinates?

anonymous · Answer

radius is a we have to find the limits first i guess for this circle nd in solutione they hav taken a strip parallel to y axis and then find the limits

turingtest · Answer

you are making it harder than it is, draw the picture...

turingtest · Answer

|dw:1349880008378:dw|it's a circle of radius a, so the x and y intercepts are a.
but doing this integral in rectangular coordinates would be very nasty, so we should convert everything to polar coordinates.
Do you know how to convert x, y, and dA to polar coordinates?
(we will find the bounds after)

anonymous · Answer

|dw:1349880050613:dw|

anonymous · Answer

can you tell me how and why this strip is taken?

turingtest · Answer

Does the problem say you $have$ to do this in rectangular coordinates?

anonymous · Answer

no

turingtest · Answer

Then forget the strip and stop thinking in rectangular coordinates. Change them to $polar$. Do you know what I mean at all?

anonymous · Answer

y has limits from 0 to $$\sqrt{a^2-x^2}$$

anonymous · Answer

and x has limits from 0 to a and then it is integrated but i dnt know why at all

turingtest · Answer

polar polar polar
you are not listening to me

anonymous · Answer

its all done on this strip basis.wtf!!

turingtest · Answer

you are writing in rectangular coordinates, change them to polar

anonymous · Answer

answer is in rectanglar cordinates

turingtest · Answer

The answer should be the same whether you use rectangular or polar coordinates

anonymous · Answer

how ?? tell me the soltn plzz

turingtest · Answer

I am not here to give you the whole solution I am here to guide you.
Let's do it in rectangular (since that seems to be what the problem requires, ugh...)
Look at the region:|dw:1349880595113:dw|for any strip that we draw the bottom of the strip will be the x-axis (y=0) and the top will be the function of the circle $y=\sqrt{a^2-x^2}$
hence the bounds on y are between those two values
$$0\le y\le\sqrt{a^2-x^2}$$do you understand that now?

anonymous · Answer

yes but for x ?

anonymous · Answer

why the value of x `is from 0 to a? u did it right i want to give u a medal but dnt know how to :p

turingtest · Answer

click "best response" (not that I care much for medals, I just like helping)

the x range comes from the same ideas that you use in calc I, look at the range on x in the first quadrant...

anonymous · Answer

i couldnt get u this time

turingtest · Answer

|dw:1349881128687:dw|x can only vary between 0 (since it's in the first quadrant) and the radius a. After that there is no more function to integrate.