(1 point) A) Use Green's theorem to compute the area inside...

(1 point) A) Use Green's theorem to compute the area inside the ellipse $frac{x^2}{17^2} + frac{y^2}{15^2} = 1$. Use the fact that the area can be written as $iint_R 1 , dy , dx = frac{1}{2} int_C -y , dx + x , dy$ where $C$ represents the ellipse. Then find a parametrization of the ellipse and use it to compute the line integral. Hint for the parametrization: $x(t) = 17 cos(t)$. The area is B) Find a parametrization of the curve $x^{2/3} + y^{2/3} = 5^{2/3}$ and use it to compute the area that it encloses. Hint for the parametrization: $x(t) = 5 cos^3(t)$.

Transcript

00:01 In this problem we are going to compute some areas defined by some curves and we are going to do that by using the greens theorem in that case.

00:16 Okay so for the first question, let's call it part a, we have this ellipse given by x squared or a square plus y squared or b squared equal to 1.

00:33 And here a is equal to 17 and b is equal to 15.

00:42 I don't want to write 17 and 15 all the time so i'm just going to call them a and b.

00:49 Okay the greens theorem tells us that the area is given by this loop integral, one half times this integral that goes over this.

01:09 Closed loop we have minus y dx plus x dy y dy i mean if we really insist on imagining or visualizing the situation we have something like this we have this x y here we have this ellipse or some elliptic shape we have a here b here now from our earlier courses of geometry and whatever those courses are.

01:54 We know that the area of an ellipse is given by pi times a times b.

02:01 So basically we are going to prove that by means of this theorem.

02:09 So let's do that.

02:12 And in order to do that, we need to parameterize this equation.

02:17 We want some parameter that goes around this loop this closed line.

02:25 So an easy way to do that is simply to use signs and cosines because we have the squares and we have this one on the other side.

02:37 So we have we can have x equal to a times cosine of some theta or you can call it also t doesn't matter and we have y equal to b times sine theta so in this case we are measuring this angle theta from the post of x -axis going in the counterclockwise direction and when this angle theta makes makes an angle two pi we are going to form and this angle will form a a closed loop and instead of using this loop integral notation we are going to say that our integral over theta is from 0 to 2 pi so that is the whole motivation whole idea of these calculations okay so this is our starting point and by the way let's check it if it makes sense.

03:45 So we have x squared, which is just a squared times cosine squared.

03:51 We divide by a squared.

03:53 So we have cosine theta squared plus we have this y squared over b squared, namely b squared times sine squared over b squared.

04:05 So we have sine theta squared is equal to one.

04:11 Everything is consistent here.

04:14 Now let's take, let's consider differentials.

04:18 We have d x equal to minus a times sine theta d theta and for y we have d y equal to b cosine theta and as i have just stated earlier we will we will replace this integral this loop integral by an integral from 0 to 2 pi over this theta variable.

05:02 Okay, let's do that.

05:06 This area is equal to 1 half from 0 to 2 pi.

05:16 Okay, so we have minus y.

05:19 So it is just minus b sine theta.

05:25 And we have this dx...

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