Math 225: Calculus III    Quiz 9    4/8&10/97

<<Calculus`Math225`

Let R be the region in the first quadrant defined by
    1 <= 2x^2 - y^2 <= 2
    1 <= 2y^2 - x^2 <= 2
Use the change of variables
    u == 2x^2 - y^2
    v == 2y^2 - x^2
to transform the integral of f[x_,y_] = x y over R into an iterated integral in u and v.  (You do not need to evaluate the integral, but you should simplify your answer.)

Solution

Here's what the region looks like (not required):

p1 = ParametricPlot[{Sqrt[(y^2+1)/2],y}, {y,0,2}];
p2 = ParametricPlot[{Sqrt[(y^2+2)/2],y}, {y,0,2}];
p3 = ParametricPlot[{x,Sqrt[(x^2+1)/2]}, {x,0,2}];
p4 = ParametricPlot[{x,Sqrt[(x^2+2)/2]}, {x,0,2}];

Show[p1,p2,p3,p4,PlotRange->{{0,2},{0,2}},AxesOrigin->{0,0},
Ticks->{{0,0.5,1,1.5,2},{0,0.5,1,1.5,2}}]

First, solve for x and y. Multiply the first equation by 2 and add it to the second:
     2u == 4x^2 - 2y^2
      v == 2y^2 - x^2
 2u + v == 3x^2
      x == Sqrt[(2u + v)/3]
Similarly,
      y == Sqrt[(2v + u)/3]
We take the positive roots since the region is in the first quadrant. We can check this with Solve[]:

Solve[ {u == 2x^2 - y^2,v == 2y^2 - x^2}, {x,y}]

The Jacobian is:

m = {{D[Sqrt[(2u + v)/3],u],D[Sqrt[(2u + v)/3],v]},
     {D[Sqrt[(2v + u)/3],u],D[Sqrt[(2v + u)/3],v]}};
m //MatrixForm

j = Det[m]

In one step:

?J

J[Sqrt[(2u + v)/3],Sqrt[(2v + u)/3],{u,v}]

The function in the new coordinates is x y -> Sqrt[(2u + v)/3] Sqrt[(2v + u)/3]. Multiplying by  the Jacobian gives:

Sqrt[(2u + v)/3] Sqrt[(2v + u)/3] * j

The new region is, of course, described by
    1 <= u <= 2
    1 <= v <= 2
so the integral of f[x,y] over R equals

Integrate[1/12, {u,1,2}, {v,1,2}]

It is worth noting that this problem can also be done easily by taking advantage of the formula:
    J[x[u,v], y[u,v], {u,v}] == 1/J[u[x,y], v[x,y], {x,y}]
This formula is not always "useful" because the right hand side is still in terms of the variables x and y. In this problem however:

m = {{D[2x^2-y^2,x],D[2x^2-y^2,y]},
     {D[2y^2-x^2,x],D[2y^2-x^2,y]}};
m //MatrixForm

j = Det[m]

Thus, the new integrand is x y (1/j) == 1/12 as above!

Let C be the curve y == x^2 from 0 <= x <= 1. Evaluate the line integral of f[x_,y_] = Sqrt[y] over the curve C.

Solution

The curve is parameterized by

r[t_] = {t, t^2};

for 0 <= t <= 1. The integral is thus equal to the integral from 0 to 1 of

Sqrt[t^2] Norm[r'[t]]

Integrate[t Sqrt[1 + 4t^2], {t,0,1}]

This is easily done by hand by making the substitution u == 1 + 4t^2, du == 8t dt.
In one step:

?LineIntegrate

LineIntegrate[ Sqrt[y], r[t], {t,0,1}]