2527 The transformation x = au, y = bv (a > 0,b > 0) canbe rewritten as x/a = u, y/b =

Chapter 14, Problem 25

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The transformation $x=a u, y=b v(a>0, b>0)$ can be rewritten as $x / a=u, y / b=v$, and hence it maps the circular region

$u^{2}+v^{2} \leq 1$

into the elliptical region

$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}} \leq 1$

$\int \int_{R} \sqrt{16 x^{2}+9 y^{2}} d A$ where $R$ is the region enclosed by the ellipse $\left(x^{2} / 9\right)+\left(y^{2} / 16=1\right)$.

Equation Transcription:

$x=au,\ y=bv(a>0,b>0)$

$x/a=u,y/b=v$

${u}^{2}+{v}^{2}\leq{}1$

$\frac{{x}^{2}}{{a}^{2}}+\frac{{y}^{2}}{{b}^{2}}\leq{}1$

$\int\limits_{\ }^{\ }\int\limits_{R}^{\ }\sqrt{{16x}^{2}+{9y}^{2}}dA$

$R$

$({x}^{2}/9)+({y}^{2}/16=1).$

Text Transcription:

x=au, y=bv(a>0,b>0)

x/a=u,y/b=v

u^2+v^2 leq 1

x^2 /a^2 + y^2 /b^2 leq 1

integral integral_R sqrt 16x^2 + 9y^2 dA

(x^2 /9) + (y^2 /16 = 1)

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