Question

Consider the transformation T: x = \frac{56}{65}u - \frac{33}{65}v, \ \ y = \frac{33}{65}u + \frac{56}{65}v
A. Computer the Jacobian:
\frac{\partial(x, y)}{\partial(u, v)} =
B. The transformation is linear, which implies that ittransforms lines into lines. Thus, it transforms the squareS:-65 \leq u \leq 65, -65 \leq v \leq 65 into a square T(S) with vertices:
T(65, 65) =
T(-65, 65) =
T(-65, -65) =
T(65, -65) =
C. Use the transformation T to evaluate the integral\int \!\! \int_{T(S)} \ x^2 + y^2 \ {dA}

Answer 1

Answer:

Step-by-step explanation:

$T: x = \frac{56}{65}u - \frac{33}{65}v, \ \ y = \frac{33}{65}u + \frac{56}{65}v$

A)

$\frac{d(x,y)}{d(u,v)} =\left|\begin{array}{ccc}x_u&x_v\\y_u&y_v\end{array}\right|$

$=(\frac{56}{65} )^2+(\frac{33}{65} )^2\\\\=\frac{(56)^2+(33)^2}{(65)^2} \\\\=\frac{4225}{4225} \\\\=1$

B )

$S:-65 \leq u \leq 65, -65 \leq v \leq 65$

$T(65,65)=(x=\frac{56}{65} (65)-\frac{33}{65} (65),\ \ y =\frac{33}{65} (65)+\frac{56}{65} (65)\\\\=(23,89)$

$T(-65,65)=(-56-33,\ \ -33+56)\\\\=(-89,23)$

$T(-65,-65) = (-56+33,-33-56)\\\\=(-23,-89)$

$T(65,-65)=(56+33, 33-56)\\\\=(89,-23)$

C)

$\int \!\! \int_{T(S)} \ x^2 + y^2 \ {dA}$

$=\int\limits^{65}_{v=-65} \int\limits^{65}_{u=-65}(x^2+y^2)(\frac{d(x,y)}{d(u,v)} du\ \ dv$

Now

$x^2+y^2=(\frac{56}{65} u-\frac{33}{65} v)^2+(\frac{33}{65} u+\frac{56}{65} v)^2$

$[(\frac{56}{65} )^2+(\frac{33}{65}) ^2]u^2+[(\frac{33}{65} )^2+(\frac{56}{65}) ^2]v^2$

$=\frac{(65)^2}{(65)^2} u^2+\frac{(65)^2}{(65)^2} v^2=u^2+v^2$

$\int \!\! \int_{T(S)} \ x^2 + y^2 \ {dA}$

$=\int\limits^{65}_{v=-65} \int\limits^{65}_{u=-65}(u^2+v^2) du\ \ dv$

$=\int\limits^{65}_{-65}\int\limits^{65}_{-65}u^2du \ \ dv+\int\limits^{65}_{-65}\int\limits^{65}_{-65}v^2du \ \ dv$

By symmetry of the region

$=4\int\limits^{65}_0 \int\limits^{65}_0u^2 du \ \ dv + u\int\limits^{65}_0 \int\limits^{65}_0v^2 du \ \ dv$

$= 4(\frac{u^3}{3} )^{65}_{0}(v)_0^{65}+(\frac{v^3}{3} )^{65}_{0}(u)_0^{65}\\\\=4[\frac{(65)^4}{3} +\frac{(65)^4}{3} ]$

$=\frac{8}{3} (65)^4$