Respuesta :
Answer:
53.7528
Step-by-step explanation:
Notice that when
[tex]xy = 1 ,\,\,\, xy = 16 , \,\,\, xy^2 = 1 \,\,\,, xy^2 = 36 \\\\[/tex]
If you set
[tex]u = xy , v = xy^2[/tex]
as they suggest, then
[tex]{\displaystyle y = \frac{v}{u}} \,\,\,\, \text{and} \,\,\,\, \\\\{\displaystyle x = \frac{u}{y} = \frac{u}{v/u} = \frac{u^2}{v} }[/tex]
Then
[tex]{\diplaystyle \frac{\partial(x,y)}{\partial(u,v)}} =\det \begin{pmatrix} 2u/v && -u^2/v^2 \\ -v/u^2 && 1/u \end{pmatrix} = \frac{1}{v} }[/tex]
Therefore
[tex]{\displaystyle \iint\limits_{D} dx\,dy = \int\limits_{1}^{36}\int\limits_{1}^{16} \frac{1}{v} \, du \, dv = 15 \ln(36) = 53.7528}[/tex]
A Jacobian matrix is formed by the first partial derivatives of a multivariate function that utilizes a training algorithm, and further calculation as follows:
Jacobian:
To evaluate the integral, cover the bounds, the integrand, and the differential area dA.
Transform the four equations in terms of u and v, notice that[tex]u= xy \ \ and \ \ xy = 1, xy = 16[/tex]
implies that [tex]1\leq u \leq 16.[/tex]
Similarly, [tex]v= xy^2\ \ and\ \ xy^2= 1 , xy^2= 25[/tex] implies that [tex]1 \leq v \leq 25[/tex]
so write this integration region as [tex]S= {(u,v) |1 \leq u \leq 18, 1 \leq v \leq 25}.[/tex]
Translate the equations from uv - plane to xy- plane. It is obtained by solving,
[tex]u= xy, y= xy^2 \\\\\left.\begin{matrix}u=xy & \\ v=xy^2& \end{matrix}\right\} \to \left.\begin{matrix}u^2=x^2y^2 & \\ v=xy^2& \end{matrix}\right\} \\\\\to x=\frac{u^2}{v}, y=\frac{v}{u}[/tex]
Convert dA part of the integral , using is [tex]dA= |\frac{\partial (x,y)}{\partial(u,v)}| dudv.[/tex]
That is,[tex]dA= \begin{vmatrix}\frac{\partial x}{\partial u} & \frac{\partial x}{\partial v}\\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{vmatrix} \ du dv \\\\[/tex]
Sampule the partial derivatives to find the Jacobian.
[tex]dA=\begin{vmatrix}\frac{2u}{v} &-\frac{u^2}{v^2} \\ -\frac{v}{u^2} &\frac{1}{u} \end{vmatrix} \ dudv\\\\=[(\frac{2u}{v}) (\frac{1}{u}) -(- \frac{u^2}{v^2})(-\frac{v}{u^2})]\ du dv\\\\=(\frac{2}{v}- \frac{1}{v}) \ dudv\\\\=\frac{1}{v}\ du dv\\\\[/tex]
The Jacobian the transformation is [tex]dA= \frac{1}{v}dudv[/tex]
The region is [tex]S={(u,v) |1\leq u \leq 16, 1\leq v\leq 25}.[/tex]
Rewrite the integral, using the transformation: [tex]S,\ x=\frac{u^2}{v} =, y=\frac{v}{u} \ \ and\ \ dA=\frac{1}{v} dudv\\\\\int\int_R 1dA =\int \int_S \frac{1}{v}\ dudv= \int^{25}_{1} \int^{16}_{1} \ \frac{1}{v} \ dudv\\\\[/tex]
Evaluate the inner integral with respect to u.
[tex]\to \int\int_R 1dA = \int^{25}_{1} \int^{16}_{1} \ \frac{1}{v} \ dudv\\\\[/tex]
by solving the value we get
[tex]= 30 \ ln (5) \approx 48.28[/tex]
Find out more about the Jacobians here:
brainly.com/question/9381576
