00:01
In this problem, we want to identify the surface whose equation is given by the polar equation r equal to 2 times sine theta.
00:09
So we want to reconvert our polar equation to cartesian.
00:21
And we do this by using the following transformation.
00:25
X is equal to r cos theta, and y is equal to r sine theta.
00:33
We then find that x is equal to 2 times cos theta sine theta, and y is equal to 2 times sine theta squared.
00:55
So here we need to use some trigonomic identities.
01:03
Two times the sine of cos theta sine theta is in fact simply the sine of 2 theta.
01:22
Additionally, 2 times the sine squared of 2 theta is also precisely 1 minus the cos of 2 theta.
01:57
And so now what we have here more clearly is the equation of a circle.
02:01
Well, to make it more clear, let's rewrite these equations as follows.
02:07
So x can't remain the same, the sine of 2 theta, but y, let's write it as 1 minus y is equal to the cos of 2 theta.
02:25
So this is an equation of a circle that is displaced by one unit on the y -axis.
02:35
And this circle has a radius equal to what? because we have ones in front of the sine and cos theta.
02:55
And for theta ranging between 0 and 2 theta, we go around the circle twice.
03:02
So let's plot our circle.
03:17
So our circle will be be displaced by one unit upwards towards the positive y -axis...