Consider the region R in the xy-plane bounded by the ellipse $\frac{x^2}{16} + \frac{y^2}{4}$=1 transformation $x = 4 u$ and $y = 2 v$. Find the area of the ellipse. Round your answer to two decimal places.
A) $25.13$
B) $8.00$
C) $16.00$
D) $50.27$
E) $201.06$

Question

Consider the region R in the xy-plane bounded by the ellipse $\frac{x^2}{16} + \frac{y^2}{4}$=1 transformation $x = 4 u$ and $y = 2 v$. Find the area of the ellipse. Round your answer to two decimal places. 
A) $25.13$
B) $8.00$
C) $16.00$
D) $50.27$
E) $201.06$

Quizplus · Accepted Answer

The area of the ellipse is given by the integral of the area of the ellipse in the uv-plane, which is $\pi(4u)^2(2v)^2 = 16\pi u^2 v^2$. Substituting $u = \frac{x}{4}$ and $v = \frac{y}{2}$, we get the area of the ellipse in the xy-plane as $\int_{-2}^{2}\int_{-1}^{1} 16\pi \left(\frac{x}{4}\right)^2 \left(\frac{y}{2}\right)^2 dy dx = 16\pi \int_{-2}^{2}\int_{-1}^{1} \frac{x^2y^2}{16} dy dx = 16\pi \int_{-2}^{2} \frac{x^2}{16} \left[\frac{y^3}{3}\right]_{-1}^{1} dx = 16\pi \int_{-2}^{2} \frac{x^2}{48} dx = 16\pi \left[\frac{x^3}{144}\right]_{-2}^{2} = 16\pi \left(\frac{8}{144} - \frac{-8}{144}\right) = \frac{16\pi}{9} \approx 25.13$.

Consider the Region R in the Xy-Plane Bounded by the Ellipse