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# The parabolas y=x^2, y=x^2+1, y=(x-2)^2, y=(x-2)^2+1

The parabolas y=x^2, y=x^2+1, y=(x-2)^2, y=(x-2)^2+1 Intersectto form a curvilinear quadrilateral R. The change of variable u=y-x^2, v=y-(x-2)^2 map R onto a square in the uv-plane. Use the jacobian of the inverse transformation to compute the area of R.

Check to make sure the question is correct. currently all parabolas are facing up, at least one should be reversed (-x or -x2), otherwise there is no upper bound, no shape and no area to compute. (1≤y<infinity)
Did you graph the four parabolas?

### 1 Answer by Expert Tutors

Al P. | Online Mathematics tutorOnline Mathematics tutor
1

The parabolas do all point up, but they also enclose a shape R that approximates a parallelogram with bounds in (x,y):    (3/4,25/16), (1,2), (1,1), and (5/4,25/16):

From T (x,y) —> (u,v):  {y-x2, y-(x-2)2}

The square is  u= y-x= 0   (for the boundary corresponding to y=x2)
u=y-(x2) = 1  (for the boundary corresponding to y=x2+1)
v = y-(x-2)2 = 0   (for the boundary corresponding to y=(x-2)2)
v = y-(x-2)2 = 1  (for the boundary corresponding to y=(x-2)2+1)

Solving T for x(u,v), y(u,v)  (i.e. writing x and y in terms of u and v):
x = (1/4)(v-u+4)
y = u + (1/16)(v-u+4)2

∂x/∂u = -1/4
∂x/∂v  = 1/4
∂y/∂u  = 1+(1/16)(-2v+2u-8)
∂y/∂v = (1/16)(2v-2u+8)

Jx,y =  (∂x/∂u)(∂y/∂v) - (∂x/∂v)(∂y/∂u)
Jx,y = -(1/4)[(1/16)(2v-2u+8)] - (1/4)[(1 + (1/16)(-2v+2u-8)]
= (-1/64)[2v-2u+8] - 1/4 - (1/64)[-2v+2u-8]

which reduces to -1/4,  so

| Jx,y | = 1/4   This is the area because we're dealing with a unit square in (u,v)

Check:
Basically, we can integrate the "hard way" in (x,y) to check the result from above.

Integrate from x=3/4 to x=1   f(x) = (x2+1) - (x-2)
and add the integral of  x=1 to x= 5/4  of  g(x) = (x-2)2+1- x2

to get   [ 2-3 - (18/16 - 36/16) ]  + [ -50/16 + 100/16 - (-2 + 5) ]
=    -1   +18/16  +  50/16 - 3
=     -4  + 68/16
=     4/16
=  1/4   (looks ok)