The problem: $\int\int_A x\ dxdy$, where A is the area delimited by the parabolas $y = x^2, y = x^2 + 4, y = (x-1)^2, y = (x-1)^2 + 4$.

I figured this should be easy but somehow I'm stuck. What I have done:

Using the limits I construct these inequalities: $0 \leq y - x^2 \leq 4$ and $0 \leq y - (x-1)^2 \leq 4$. So I change variables to $u = y - x^2,\ v = y- (x-1)^2$. Using inverse function theorem, I calculate $\frac{\partial(u, v)}{\partial(x,y)}$, invert it and take the absolute value, which yields $\frac{1}{2}$: $\frac{\partial u}{\partial x} = -2x, \frac{\partial u}{\partial y} = 1, \frac{\partial v}{\partial x} = -2x + 2, \frac{\partial v}{\partial y} = 1 \implies -2x - (-2x+2)$.

Then I find $x(u,v)$. $u-v = y-x^2 - y + (x-1)^2 = -2x + 1 \implies x = \frac{1}{2}(1-u+v)$. The integral then becomes $\int_0^4\int_0^4 \frac{1}{4}(1-u+v)\ dudv = 4$, which WolframAlpha confirms. My book tells me the answer should be 20.

So I must have stepped wrong somewhere in the change of variables or setting up my integral... I just don't see where and what..?


Here is the region A :

Graph of functions delimiting region A

I think you are correct.

With slightly other notation, I define $\Phi : [0,4]^2 \rightarrow A,\: (u,v) \mapsto (x,y)$ and find that $\det J_{\Phi}=-\frac{1}{2}$. Then,

$$\int \int_A x~dx~dy = \int_0^4 \int_0^4 \frac{1}{2}(1-u+v)~\frac{1}{2}~du~dv$$

What is your book ?

  • $\begingroup$ My book is "Flervariabel analyse med lineær algebra"(Multivariable analysis w/ linear algebra), it's norwegian. I asked a teacher at school and he thought this was the correct answer too - so maybe the book is wrong. Thank you. $\endgroup$ – Zak Laberg Mar 11 '15 at 16:40
  • $\begingroup$ @ZakLaberg Maybe the book mistook 16/4 for 16+4 ;) $\endgroup$ – bela83 Mar 12 '15 at 12:50

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