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We have the double integral:

$$\int \int_D 2x + 3y \; dx\;dy$$

The domain in which we want to calculate this is the flat region defined by the curves:

$$y = x^2 \; ; \; y=x$$

Then, through the decomposition rules we resolve the internal integral to $dy$, and to do this we find the copy ordinates of the minimum and maximum points of the domain, which are precisely

$$y = x^2 \; ; \; y=x$$

While the minimum and maximum points abiscissas will be the external integral range

$$\int_{0}^{1} dx \int^{x}_{x^2} 2x + 3y \; dy$$

The coordinates are found by solving to $y$ the curves that define the domain : for the abscissas, does there exist a mathematical method, or should we simply be intuitive?

Thank you in advance

dmtri
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user3204810
  • 299

2 Answers2

3

The set up is correct and a simple plot of the graph can help to guide in order to find the abscissa range which can then be found by the equation

$$x^2=x \iff x(x-1)=0 \iff x=0 \lor x=1$$

Note that to define the region $D$ properly we also need some other condition that is for example that the integral must be proper or that

$$D=\{(x,y)\in \mathbb{R^2}: x^2\le y\le x\}$$

user
  • 154,566
0

You can also define new variables $$u = x, \quad v = \frac{y}x$$

We see that $u = x \in [0,1]$ and $y \in [x^2, x] \iff v = \frac{y}x \in [x,1]=[u,1]$.

The Jacobian is given by

$$\frac1J = \begin{vmatrix} \partial_xu & \partial_yu \\ \partial_xv& \partial_yv\end{vmatrix} = \begin{vmatrix} 1 & 0 \\ -\frac{y}{x^2}& \frac1x\end{vmatrix} = \frac1x$$

so $|J| = |x| = x = u$.

Since $y = xv = uv$, the resulting integral is

$$\int_{u=0}^1 \int_{v=u}^1 (2u+3uv)u\,dv\,du = \frac{11}{30}$$

which is the same as $\int_{0}^{1} \int^{x}_{x^2} (2x + 3y) \,dy\,dx$.

mechanodroid
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