Puzzling Double integral

Question

I was asked to solve this double integral: Compute the area between $y=2x^2$ and $y=x^2$ and the hyperbolae $xy=1$ and $xy=2$ in

$$ \iint dx \,dy$$

I tried to solve it starting with considering that

$$x^2 \leq y \leq 2x^2 $$

suitabile for integration interval in $y$, obtaining the incomplete form

$$ \int^{x^2}_{2x^2} \int_\ldots^\ldots dx \,dy$$

but I also have $$1 \leq xy \leq 2$$ and I would obtain a result in which I still have one independent variabile.

Please, can anyone help me? Thanks in advance.

Answer 1

HINT

Let consider the change of variables

• $u=x^2\implies 1\le u\le 2$
• $v=xy\implies 1\le v\le 2$

and

$$dudv=|J|dxdy=\begin{vmatrix}2x&0\\y&x\end{vmatrix}dxdy=2x^2dxdy\implies dxdy=\frac1{2u}dudv$$

Answer 2

Plot the four functions. See where they intersect. Once you do, you will find that you can rewrite this integral as:

$$\displaystyle \int_{ \tfrac{1}{ \sqrt[3]{2} } }^1\left( 2x^2-\dfrac{1}{x}\right)dx + \int_1^{\sqrt[3]{2}}\left(\dfrac{2}{x}-x^2\right)dx$$

Answer 3

$y \in [x^2, 2x^2]$ can be interpreted as $\frac{y}{x^2} \in [1,2].$

Consider the change of variables

• $u = \frac{y}{x^2} \in [1,2]$

• $v = xy \in [1,2]$

The Jacobian is given by

$$\frac1J = \begin{vmatrix} u_x & u_y \\ v_x & v_y \end{vmatrix} = \begin{vmatrix} -\frac{2y}{x^3} & \frac1{x^2} \\ y & x \end{vmatrix} = -\frac{3y}{x^3} = -3u$$

so $$dx\,dy = |J| \,du\,dv = \frac1{3u} \, du\,dv$$

Therefore you need to calculate

$$\int_{[1,2]^2} \frac{1}{3u}\,du\,dv= \frac13\ln 2$$