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Answer:
96π
Step-by-step explanation:
The transformation x = au , y = bv (a>0, b>0) can be re-written as;
[tex]\dfrac{x}{a}= u , \dfrac{y}{b}=v ;[/tex] which maps the circular region u² + v² ≤ 1 into the elliptical region
[tex]\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}\le 1[/tex]
Integrating and transforming the elliptical region into a circular region;
[tex]\int \int _R \sqrt{16x^2+9y^2}\ dA[/tex]
here;
R = region enclosed by the ellipse.
[tex]\dfrac{x^2}{9}+\dfrac{y^2}{16}=1[/tex]
Let x = 3u ; y = 4v
Jocobian = J(x,y)
[tex]= \left[\begin{array}{ccc}\dfrac{\partial x}{\partial u}&\dfrac{\partial x}{\partial v}\\\dfrac{\partial y}{\partial u}&\dfrac{\partial y}{\partial v}\\\end{array}\right] \\ \\ \\ \\ = \left[\begin{array}{ccc}3&0\\0&4\\\end{array}\right] \\ \\ \\ = 12[/tex]
So;
[tex]\int \int_R \sqrt{16x^2+9y^2 \ dA}= \int \int _D\sqrt{16(4u)^2+9(3v)^2} \ |J| dudv \\ \\ \\= 12 \int \int _D \sqrt{x^2 +v^2} \ \ (12) dudv \\ \\ \\ =144 \int \int _D \sqrt{x^2 +v^2} \ \ dudv[/tex]
Using polar coordinates;
[tex]u = rcos \theta ; v = rsin \theta \\ \\ \implies u^2+v^2 = r^2 ; dudv = rdr \theta \\ \\ limits :\\ \\ 0 \le r \le 1, 0\le \theta \le 2 \pi[/tex]
∴
[tex]\int \int _R \sqrt{16x^2 + 9y^2 }\ dA[/tex]
[tex]= 144 \int ^{2 \pi}_{0} \int ^{1}_{0} \ r (rdrd \theta) \\ \\ = 144 \int ^{2 \pi}_{0} \int ^{1}_{0} \ r^2drd \theta \\ \\ = 144 \int ^{2 \pi}_{0}(\dfrac{r^3}{3})^1_0 r d \theta \\ \\ = 144 ( \dfrac{1}{3}) \int ^{2 \pi}_{0} \ d \theta \\ \\ = 144(\dfrac{1}{3}) ( 2 \pi) \\ \\ \mathbf{= 96 \pi}[/tex]