anonymous
  • anonymous
Find the area of the domain D which is in the first quadrant and bounded by xy=1, xy=3, y=x, y=3x.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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zzr0ck3r
  • zzr0ck3r
solve for y, and draw a picture with all the curves...
anonymous
  • anonymous
Is there a way to do this without sketching?
zzr0ck3r
  • zzr0ck3r
unless you have a great mind for imagining how a closed bouded region will look like....I dont

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zzr0ck3r
  • zzr0ck3r
bounded
anonymous
  • anonymous
|dw:1370719868456:dw| Let's use a change of variables \(u=xy,v=\frac{y}x\) to get new bounds $$u=1,u=3,v=1,v=3$$|dw:1370720068646:dw|Now, we compute our Jacobian:$$\det\begin{bmatrix}u_x&u_y\\v_x&v_y\end{bmatrix}=\det\begin{bmatrix}y&x\\-\frac{y}{x^2}&\frac1x\end{bmatrix}=\frac{y}x+\frac{xy}{x^2}=2\frac{y}{x}=2v$$ Now throw it all together:$$\iint_DdA=\int_1^3\int_1^3(2v)\,du\,dv=\int_1^32v\int_1^3du\,dv=4\int_1^3dv=8$$
anonymous
  • anonymous
Becareful, however, since our transformed region (pullback) is also mapped to by our third quadrant where our functions also bound an identical area:|dw:1370721047152:dw|
anonymous
  • anonymous
So halve our result to get \(8/2=4\).

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