Respuesta :
Answer:
The area can be written as
[tex]\int\limits_1^2 \int\limits_1^3 u(\frac{1}{v} - v \, ln(v)) \, du \, dv = 0.2274[/tex]
And the value of it is approximately 1.8117
Step-by-step explanation:
x = u/v
y = uv
Lets analyze the lines bordering R replacing x and y by their respective expressions with u and v.
- x*y = u/v * uv = u², therefore, x*y = 1 when u² = 1. Also x*y = 9 if and only if u² = 9
- x=y only if u/v = uv, And that only holds if u = 0 or 1/v = v, and 1/v = v if and only if v² = 1. Similarly y = 4x if and only if 4u/v = uv if and only if v² = 4
Therefore, u² should range between 1 and 9 and v² ranges between 1 and 4. This means that u is between 1 and 3 and v is between 1 and 2 (we are not taking negative values).
Lets compute the partial derivates of x and y over u and v
[tex] x_u = 1/v [/tex]
[tex] x_v = u*ln(v) [/tex]
[tex] y_u = v [/tex]
[tex] y_v = u [/tex]
Therefore, the Jacobian matrix is
[tex]\left[\begin{array}{ccc}\frac{1}{v}&u \, ln(v)\\v&u\end{array}\right][/tex]
and its determinant is u/v - uv * ln(v) = u * (1/v - v ln(v))
In order to compute the integral, we can find primitives for u and (1/v-v ln(v)) (which can be separated in 1/v and -vln(v) ). For u it is u²/2. For 1/v it is ln(v), and for -vln(v) , we can solve it by using integration by parts:
[tex]\int -v \, ln(v) \, dv = - (\frac{v^2 \, ln(v)}{2} - \int \frac{v^2}{2v} \, dv) = \frac{v^2}{4} - \frac{v^2 \, ln(v)}{2}[/tex]
Therefore,
[tex]\int\limits_1^2 \int\limits_1^3 u(\frac{1}{v} - v \, ln(v)) \, du \, dv = \int\limits_1^2 (\frac{1}{v} - v \, ln(v) ) (\frac{u^2}{2}\, |_{u=1}^{u=3}) \, dv= \\4* \int\limits_1^2 (\frac{1}{v} - v\,ln(v)) \, dv = 4*(ln(v) + \frac{v^2}{4} - \frac{v^2\,ln(v)}{2} \, |_{v=1}^{v=2}) = 0.2274[/tex]
