**Transcription for Educational Website:**---### Topic: Calculating Areas Using Green's Theorem and Parametrization**A)**Use Green's theorem to compute the area inside the ellipse defined by:\[\frac{x^2}{19^2} + \frac{y^2}{14^2} = 1.\]Green's theorem states that the area can be represented as:\[\iint_D dx \, dy = \frac{1}{2} \oint_{\partial D} -y \, dx + x \, dy.\]**Hint:** Use the parametrization \(x(t) = 19 \cos(t)\).**Result:** The area of the ellipse is **836**.---**B)**To find the area of the interior defined by the curve:\[x^{2/3} + y^{2/3} = 9^{2/3},\]find a suitable parametrization of the curve.**Hint:** Use the parametrization \(x(t) = 9 \cos^3(t)\).**Result:** The area is \(\frac{81\pi}{4}\).---These exercises illustrate techniques for calculating areas enclosed by curves using calculus tools like Green's Theorem and curve parametrization.

(A). Given, x2192+y2142=1, use the fact that the area can be written as: ∫∫Ddxdy=12∫∂D-ydx+xdy.Now the parametric equation of the ellipse x2192+y2142=1 is x(t)=19cos(t), y(t)=14sin(t), 0≤t≤2π. Therefore, Area=∬Ddxdy, =12∫∂D-ydx+xdy (By Green's theorem) =12∫02π-14sin(t)·(-19sin(t)+19cos(t)·14cos(t)dt=2662∫02πsin2(t)+cos2(t)dt,=133∫02π(1)dt ∵ sin2θ+cos2θ=1=133∫02πdt, =133t02π, =133(2π)≈836. Hence the required area is 836 . (B). Given the curve x23+y23=923 . Now the parametric equation of the given curve is x(t)=9cos3(t), & y(t)=9sin3(t), 0≤t≤2π. Therefore, Area=∬Ddxdy, =12∫∂D-ydx+xdy (By Green's theorem) =12∫02π-9sin3(t)·(-27cos2(t)sin(t))+9cos3(t)·(27sin2(t)cos(t))dt=2432∫02πsin4(t)cos2(t)+cos4(t)sin2(t)dt,=2432∫02πsin2(t)cos2(t)sin2(t)+cos2(t)dt,=2438∫02π4sin2(t)cos2(t)(1)dt ∵ sin2θ+cos2θ=1=2438∫02π2sin(t)cos(t)2dt, =2438∫02πsin2(2t)dt, =2438∫02π121-cos(4t)dt, =24316∫02π1-cos(4t)dt, =24316t-sin(4t402π, =243162π, =243π8. Hence the required area of the interior is 243π8.