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-x^{2}, y-(x-2)^{2}}

The square is u= y-x^{2 }= 0 (for the boundary corresponding to y=x^{2})

u=y-(x^{2}) = 1 (for the boundary corresponding to y=x^{2}+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)

J_{x,y }= (∂x/∂u)(∂y/∂v) - (∂x/∂v)(∂y/∂u)

J_{x,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

** | J**_{x,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) = (x^{2}+1) - (x-2)^{2 }

and add the integral of x=1 to x= 5/4 of g(x) = (x-2)^{2}+1- x^{2}

** **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)

** **

## Comments

^{2}), otherwise there is no upper bound, no shape and no area to compute. (1≤y<infinity)