1

Problem

I'm having trouble with the following integral:

$\int \int_D \frac{x}{y} dx dy $ where D is the area bounded by $1 \leq 2x+y \leq 5, \quad 4x \leq y \leq 8x$

My attempt at a solution

Substitution: $ u = 2x + y, \quad v= y \ \implies x = \frac{1}{2}(u-v)$

Jacobian:

$ \begin{vmatrix} \frac{1}{2} & -\frac{1}{2} \\ 0 & 1 \end{vmatrix} = \frac{1}{2}$

Limits for u:

$u: 1\to 5$

Limits for y:

$4x \leq y \implies 2(u-v) \leq v \implies \frac{2}{3}u \leq v \\ y \leq 8x \implies v \leq 4(u-v) \implies v \leq \frac{4}{5}u$

$v: \frac{2}{3}u \to \frac{4}{5}u$

Everything put together:

$\int \int_D \frac{x}{y} dx dy = \int \int_D \frac{\frac{1}{2}(u-v)}{v} \cdot \frac{1}{2} dudv = \int_1^5 \int_{\frac{2}{3}u}^{\frac{4}{5}u} \Big(\frac{u}{4v}-\frac{1}{4} \Big)dv du$

I put this into wolframs double integral calculator and get:

$3 \log(\frac{6}{5})-\frac{2}{5}$ which is wrong. Am I doing something wrong?

Mattias Johnson
  • 19
  • 3

1 Answers1

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Hint. Integrate w.r.t. $y$ immediately:

$$\int_{4x}^{8x} \frac{dy}{y}=\log 2$$

Now the only thing you have left is finding the limits for $x$ from the two inequalities and integrating:

$$\log 2\int_{x_1}^{x_2} xdx$$

Note that the first integration is equivalent to a change of variable:

$$y=x t$$

$$\int_{4x}^{8x} \frac{dy}{y}=\int_{4}^{8} \frac{dt}{t}$$

Yuriy S
  • 31,474