00:01
Hi, so in this question we're going to be solving our given equation in double, using double integrals, but first we have to translate all of these coordinates into polar coordinates.
00:17
So we have three equations that i've written that will help us do this.
00:23
So first let's start trying to solve for our r.
00:27
So so we have that our x is our first part.
00:34
So we're going to say that 0 is equal to our cos theta.
00:47
So we're going to say that we can divide by our, excuse me, our cos theta.
01:04
And we're going to have 0 is equal to r.
01:09
And we also have our equation the square root of 1 minus y squared so we're going to have that as being equal to our x so 1 minus y squared is equal to r cos theta so what we're going to do from here is we're going to square both of our sides and we're going to have one of our sides and we're going to have one of minus y squared is equal to r squared cos squared theta and we're going to have one minus now we know r y is equal to r sign so we're going to have r squared sine squared is equal to cosine squared squared r squared so we can write one is equal to r squared sine squared plus r squared cose squared theta so we have one is equal to r squared sine squared theta plus cose squared theta and this is equal to one so we're going to have r is equal to plus minus one and so now we're going to solve for our theta values so we have that our y is equal to one and zero and so we can plug that into our y equation so we'll have one is equal to sine theta, r theta, sorry.
03:36
And we have 0 is equal to r sine theta.
03:44
So you have 0 is equal to sine theta, and we know that that is true when, so we'll do the sine of to the negative one of 0 is equal to we know that is true when sine is equal to when theta, sorry, is equal to pi over two.
04:10
So we have pi over two is equal to theta...