Change of Variables Calculator

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Example

Created on 2024-06-20Asked by Emma Rivera (Solvelet student)

Change the variables in the double integral

\iint_R x^2 y \, dx \, dy

using

u = x + y

and

v = x - y

Solution

Step 1: Identify the new variables

u = x + y

and

v = x - y

. Step 2: Find the Jacobian determinant:

J = \frac{\partial (x,y)}{\partial (u,v)} = \begin{vmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{vmatrix}.

Step 3: Solve for

x

and

y

in terms of

u

and

v

x = \frac{u+v}{2}, \quad y = \frac{u-v}{2}.

Step 4: Calculate the partial derivatives and the Jacobian determinant:

\frac{\partial x}{\partial u} = \frac{1}{2}, \quad \frac{\partial x}{\partial v} = \frac{1}{2}, \quad \frac{\partial y}{\partial u} = \frac{1}{2}, \quad \frac{\partial y}{\partial v} = -\frac{1}{2},

J = \begin{vmatrix} \frac{1}{2} & \frac{1}{2} \ \frac{1}{2} & -\frac{1}{2} \end{vmatrix} = \left( \frac{1}{2} \cdot -\frac{1}{2} \right) - \left( \frac{1}{2} \cdot \frac{1}{2} \right) = -\frac{1}{4} - \frac{1}{4} = -\frac{1}{2}.

Step 5: Substitute and simplify the integrand:

x^2 y = \left( \frac{u+v}{2} \right)^2 \left( \frac{u-v}{2} \right).

Step 6: Set up the new integral with the Jacobian:

\iint_R x^2 y \, dx \, dy = \iint_{R'} \left( \frac{(u+v)^2 (u-v)}{8} \right) \left( \frac{1}{2} \right) \, du \, dv.

Step 7: Conclusion. The new integral in terms of

u

and

v

is:

\iint_{R'} \frac{(u+v)^2 (u-v)}{16} \, du \, dv.

Solved on Solvelet with Basic AI Model

Some of the related questions asked by Eleanor Rivera on Solvelet

1. Evaluate the integral

\iint_D (x^2 + y^2) \, dA

over the region

D

bounded by the curves

x = y^2

and

x = 2y

2. Use change of variables to evaluate the integral

\iint_{R} (x^2 + y^2) \, dA

over the region

R

bounded by the ellipse

\frac{x^2}{4} + \frac{y^2}{9} = 1

DefinitionMainly in calculus, differential equations and probability theory to convert integrals from one system of coordinates to another. Integration by substitution hence transforms the region of integration and the function to be integrated, allowing for an easier integration. It is also applied to the solution of differential equations, transforming them into canonical forms that are simple to solve. For example, change of variables to polar coordinates (x = r cos θ, y = r sin θ) reduces the basic area element in a special case (r112 + r113 = 0 over r = 0 to r in this example):

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