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

Explanation:

`T: x = (56)/(65)u - (33)/(65)v, \ \ y = (33)/(65)u + (56)/(65)v`

A)

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

`=((56)/(65) )^2+((33)/(65) )^2\\\\=((56)^2+(33)^2)/((65)^2) \\\\=(4225)/(4225) \\\\=1`

B )

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

`T(65,65)=(x=(56)/(65) (65)-(33)/(65) (65),\ \ y =(33)/(65) (65)+(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)((d(x,y))/(d(u,v)) du\ \ dv`

Now

`x^2+y^2=((56)/(65) u-(33)/(65) v)^2+((33)/(65) u+(56)/(65) v)^2`

`[((56)/(65) )^2+((33)/(65)) ^2]u^2+[((33)/(65) )^2+((56)/(65)) ^2]v^2`

`=((65)^2)/((65)^2) u^2+((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((u^3)/(3) )^(65)_(0)(v)_0^(65)+((v^3)/(3) )^(65)_(0)(u)_0^(65)\\\\=4[((65)^4)/(3) +((65)^4)/(3) ]`

`=(8)/(3) (65)^4`