- #1
Emspak
- 243
- 1
Homework Statement
Let B [itex]\in[/itex] ℝ2
the region is bounded by [itex]x^2 + y^2 = 4, \ x^2 + y^2 = 1, \ x^2 = y, \ 2x^2 = y[/itex]
evaluate
[tex]\iint \limits_{B}\frac {2x^2+y^2}{xy}[/tex]The attempt at a solution
I needed to get the limits of integration. I used the following formula to start with:
[tex]\iint \limits_{B}f \circ g = \iint \limits_{B}(f \circ g) J (f \circ g) [/tex]
So I let [itex]s = y-x^2, \ t=1-x^2 - y^2[/itex] as those are easy to work with at the start. I do the following:
[itex]\frac{\partial (s,t)}{\partial (x,y)}= \begin{vmatrix}
-2x & 1\\
-2x & -2y\\
\end{vmatrix}[/itex]
The determinant gets me [itex]4xy +2x[/itex] and the inverse of that is [itex]\frac {1}{4xy +2x}[/itex]
So I put it all together thus:
[tex]\iint \limits_{B}\frac {2x^2+y^2}{xy}\frac {1}{4xy +2x} = \iint \limits_{B}\frac {2x^2+y^2}{4x^2y^2+2x^2y}[/tex]
To get the integration limits I look at where the various curves meet. In the case of y=x2 and x2+y2=1 it's at (-1/√2, 1/√2) and the parabola meets the other circle at (-√2, √2). The y=2x2 curve meets the circles at (-1/√3, √2/√3) and (-2/√3, 2√2/√3).
so the limits of x are -1/√3 to -√2 and y is from 1/√2 to 2√2/√3.
That gives me
[tex]\int\limits_{-1/\sqrt{3}}^{-\sqrt{2}}\int\limits_{1/\sqrt{2}}^{2\sqrt{2}/\sqrt{3}} \frac {2x^2+y^2}{4x^2y^2+2x^2y}\,dy\,dx[/tex]
So my question is, what I did wrong here :-) since odds are there is something.
Thanks in advance.