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Compute $$\iint_D \frac{1}{(1+(x+2y)^2)^2} \,dx\,dy$$ where $D$ is given by $x \geq 0 , \, y \geq 0, \, 1 \leq x+2y \leq 2 \\$.

I am supposed to solve it with the help of contour lines. By drawing $D$ we get a trapezoid with the corners $(1,0), \, (2,0), \, (0, 1/2), \, (0,1)$. A contour line $\gamma$ to $f(x,y) = x+2y$ would be a line segment with end points $(\lambda, \frac{\lambda}{2})$. Consider the area of a trapezoid bounded by the line segment and the line segment with corners $(0, \frac{1}{2}), \, (1,0)$: $$ A(\lambda) = \frac{1}{2}(\sqrt{(1/2)^2+1}+\sqrt{\lambda + (\lambda /2)^2)})\sqrt{(\lambda-1)^2+(\lambda/2-1/2)^2} = \frac{1}{2} \sqrt{5/4} (\lambda +1)(\lambda -1)\sqrt{5/4}=\frac{5}{8} (\lambda^2-1) $$ where, on the right side of the first equality, we are taking the mean of the sides of the trapezoid times the height. The derivative is $A'(\lambda) = \frac{5}{4}\lambda$. The double integral can be computed as a single integral: $$\int_{\lambda=1}^{\lambda=2} \frac{1}{(1+\lambda^2)^2} \frac{5}{4} \lambda \, d \lambda =\cdots=3/16$$ This answer is wrong. The answer sheet says it is $3/40$. What have I done wrong?

Michael Hardy
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Lozansky
  • 1,035

1 Answers1

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The height of the trapezoid is incorrect. Consider the line $y=2x$ which is perpendicular to the contour lines. This line intersects the contour $f(x,y) = \lambda$ at $\left(\frac{\lambda}{5} , \frac{2\lambda}{5}\right)$. So, the height is given by $$\sqrt{\left(\frac{\lambda}{5} - \frac{1}{5}\right)^2 + \left( \frac{2\lambda}{5} - \frac{2}{5}\right)^2} = \frac{1}{\sqrt{5}} (\lambda - 1).$$ Using this instead gives $A(\lambda) = \frac{1}{4} \left(\lambda^2-1\right)$ and so $A'(\lambda) = \frac{\lambda}{2}$. Hence, $$\int_{1}^{2} \frac{1}{\left(1+\lambda^2\right)^2} \frac{\lambda}{2} \: d \lambda = \frac{3}{40}.$$

user197427
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