# Transformations in double integrals. (Jacobian)

• VinnyCee
A change of variables can be useful in many situations, for example, when solving a difficult integral or when trying to simplify the integrand. In this specific problem, the transformation was used to simplify the integrand and make it easier to integrate. This technique is commonly used in multivariable calculus and can be applied to various problems involving multiple integrals.
VinnyCee
Evaluate

$$\int\int_{R} \left(2x^2 - xy - y^2\right) dx\;dy$$

by applying the transformation

$$u = x - y , v = 2x + y$$

for the region R in the first quadrant bounded by the lines

$$y = -2x + 4, y = -2x + 7, y = x - 2, y = x + 1$$

I don't even know where to start! Please help.

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I have this so far

$$y = x - u$$

$$v = 2x + x - u$$

$$v = 3x - u$$

$$3x = v + u$$

$$x = \frac{1}{3} \left(u + v\right)$$

$$y = \frac{1}{3} \left(-2u + v\right)$$

then, using partial derivatives to get the Jacobian to be $$\frac{1}{3}$$, right?

Then I use those equations and the line equations to solve for the limits of u and v:

$$u = -1\;to\;2$$

$$v = 4\;to\;7$$

The resulting transformed integral (with Jacobian in front)would then be:

$$\frac{1}{3}\int_{-1}^{2}\int_{4}^{7} v\;u\;dv\;du = \frac{33}{4}$$

Is this correct?

VinnyCee said:
$$\frac{1}{3}\int_{-1}^{2}\int_{4}^{7} v\;u\;dv\;du = \frac{33}{4}$$

Is this correct?

I think you need to transform the integrand in terms of u and v, that is:

$$\iint_R F(x,y)dxdy=\pm\iint_S F(f(u,v),g(u,v))\frac{\partial (x,y)}{\partial (u,v)}dudv$$

Note the plus/minus sign.

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Is this part correct?

$$\frac{\partial (x,y)}{\partial (u,v)} = \frac{1}{3}$$

But what is this equal to, how do I find it?

$$F(f(u,v),g(u,v)) =\;?$$

The Jacobian is correct. It's just a 2x2 determinant (1/6+1/6) right?

To transform the integrand, just substitute the expressions for x and y in terms of u and v. You know, x=1/3(u+v) and y=1/3(v-2u). So $x^2$ is . . . xy is . . . and so on. Then integrate over the transformed area in the u-v plane which is a nice square.

After your advice, i get the following

$$\frac{1}{3}\;\int_{-1}^{2}\int_{4}^{7} \left(-\frac{1}{9}\;u^2 - \frac{1}{9}\;v^2 + \frac{7}{9}\;u\;v\right)\;dv\;du$$

Is this right?

Well, I didn't simplify it but for:

$$2x^2-xy-y^2$$

with:

$$x=\frac{1}{3}(u+v)$$

$$y=\frac{1}{3}(v-2u)$$

Isn't that just:

$$2[\frac{1}{3}(u+v)]^2-\frac{1}{9}(u+v)(v-2u)-\frac{1}{9}(v-2u)^2$$

Simplified and hopefully I am understanding the concept now!

This

$$2[\frac{1}{3}(u+v)]^2-\frac{1}{9}(u+v)(v-2u)-\frac{1}{9}(v-2u)^2$$

simplifies to

$$-\frac{4}{9}\;u^2 + \frac{2}{9}\;v^2 + \frac{7}{9}\;u\;v$$

(I think )

Now that the Jacobian has been checked and we just simplified the re-substitution (correct term? ), are the limits of integration correct?

$$\frac{1}{3}\;\int_{-1}^{2}\int_{4}^{7} \left(-\frac{4}{9}\;u^2 + \frac{2}{9}\;v^2 + \frac{7}{9}\;u\;v\right)\;dv\;du$$

Does everything seem to be in order?

VinnyCee said:
This

$$2[\frac{1}{3}(u+v)]^2-\frac{1}{9}(u+v)(v-2u)-\frac{1}{9}(v-2u)^2$$

simplifies to

$$-\frac{4}{9}\;u^2 + \frac{2}{9}\;v^2 + \frac{7}{9}\;u\;v$$

(I think )

Now that the Jacobian has been checked and we just simplified the re-substitution (correct term? ), are the limits of integration correct?

$$\frac{1}{3}\;\int_{-1}^{2}\int_{4}^{7} \left(-\frac{4}{9}\;u^2 + \frac{2}{9}\;v^2 + \frac{7}{9}\;u\;v\right)\;dv\;du$$

Does everything seem to be in order?

Vinny, sorry I got to you late. Probably you've figured it out by now but I get just uv for the integrand:

$$\frac{1}{3}\;\int_{-1}^{2}\int_{4}^{7}(uv)dvdu=\frac{99}{12}$$

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just wondering

Just wondering... I'm learning the change of variables Jacobian things too right now and also having some problems as well.

Would someone tell me how to find the limits of integration for 'u' and 'v'?
Like you got:

u = -1 to 2
v = 4 to 7

How do you find those?

The area is bounded by y=-2x+4 and y=-2x+7, so that's y+2x=4 and y+2x=7, or v=4 to v=7, so v goes from 4 to 7, and the same for u. With this change of variables, you're actually fitting the area of integration to a rectangle.

yea. that makes so much sense now, putting it into a rectangle like that. thanks.

## 1. What is the purpose of using the Jacobian in double integrals?

The Jacobian is used to transform the coordinates of a double integral from one coordinate system to another. This allows for easier computation and evaluation of the integral in a different coordinate system.

## 2. How is the Jacobian calculated in double integrals?

The Jacobian is calculated by taking the determinant of the matrix of partial derivatives of the transformation function. This matrix is often referred to as the Jacobian matrix.

## 3. What is the significance of the Jacobian in multivariable calculus?

The Jacobian plays a crucial role in multivariable calculus as it is used to transform integrals in different coordinate systems, which is essential in solving problems involving multiple variables. It is also used in computing surface area and volume in higher dimensions.

## 4. Can the Jacobian be negative?

Yes, the Jacobian can be negative. This indicates that the orientation of the coordinate system has changed. It is essential to keep track of the sign of the Jacobian when transforming integrals to ensure the correct calculation of the integral.

## 5. Are there any other applications of the Jacobian besides double integrals?

Yes, the Jacobian has applications in other fields such as physics, engineering, and computer science. It is also used in solving differential equations, optimization problems, and in the study of vector fields and transformations in linear algebra.

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