Problem 24
Question
In Problems 23-26, evaluate the given double integral by means of an appropriate change of variables. \(\int_{-2}^{0} \int_{0}^{x+2} e^{y^{2}-2 x y+x^{2}} d y d x\)
Step-by-Step Solution
Verified Answer
Use a substitution \( u = y - x, v = x + y \) with Jacobian \( \frac{1}{2} \) to simplify and evaluate the integral.
1Step 1: Analyze the Integral
The given integral is \( \int_{-2}^{0} \int_{0}^{x+2} e^{y^{2}-2xy+x^{2}} dy\ dx \). Realize that the exponent \( y^2 - 2xy + x^2 \) can be rewritten in a different form by completing the square. This suggests that a change of variables might simplify the process.
2Step 2: Choose an Appropriate Substitution
Since the expression \( y^2 - 2xy + x^2 \) resembles \((y-x)^2\), let us choose new variables: \( u = y-x \) and \( v = x+y \). This substitution simplifies the expression. We should also calculate the Jacobian for the transformation.
3Step 3: Calculate the Jacobian
The transformation equations are \( u = y-x \) and \( v = x+y \). Then, \( x = \frac{v-u}{2} \) and \( y = \frac{v+u}{2} \). The Jacobian of the transformation is found by calculating the determinant: \[ J = \begin{vmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{vmatrix} = \begin{vmatrix} -\frac{1}{2} & \frac{1}{2} \ \frac{1}{2} & \frac{1}{2} \end{vmatrix} = \frac{1}{2} \]
4Step 4: Determine New Limits of Integration
Transform the limits of integration according to the new variables. The lines \( y=0 \) and \( y=x+2 \) in the region are now \( u=-x \) and \( u=2-x \) respectively. For \( x \) from \(-2\) to \(0\), transforming endpoints results in \( v \) going from \(-2\) to \(2\).
5Step 5: Rewrite the Integral with New Variables
The transformed integral becomes \[ \int_{-2}^{2} \int_{-v}^{v} e^{u^2} \left( \frac{1}{2} \right) dudv \]. The \( e^{u^2} \) part clarifies that substitution correctly simplified the integrand.
6Step 6: Evaluate the Integral
The integral \( \int_{-v}^{v} e^{u^2} du \) is symmetrical around \( u = 0 \), hence can be split at zero. Look for tables or mathematical software since \( e^{u^2} \) doesn't have a simple primitive function. After evaluating the integrals separately for both limits, the result reflects through symmetry, reducing evaluation challenges.
Key Concepts
Change of VariablesJacobianCompleting the SquareLimits of Integration
Change of Variables
In evaluating complex double integrals, the change of variables technique is often invaluable. The given integral initially appears complicated due to its exponent: \( y^2 - 2xy + x^2 \). By changing variables, we attempt to transform this into a simpler, more manageable form.
When using a change of variables, we typically define new variables that make the mathematical expression easier to work with. In our example, the chosen substitutions were \( u = y-x \) and \( v = x+y \). This simplification is based on recognizing patterns that fit known identities (like completing the square).
This approach benefits the computation by reducing the complexity of the integrand or by aligning the integrand with a more easily solvable form.
When using a change of variables, we typically define new variables that make the mathematical expression easier to work with. In our example, the chosen substitutions were \( u = y-x \) and \( v = x+y \). This simplification is based on recognizing patterns that fit known identities (like completing the square).
This approach benefits the computation by reducing the complexity of the integrand or by aligning the integrand with a more easily solvable form.
Jacobian
In calculus, the Jacobian matrix is a crucial tool when transforming variables in multiple integrals. It helps to account for how the change of variables affects the "volume" element of integration.
In our transformation, where \( u = y-x \) and \( v = x+y \), the Jacobian matrix, denoted \( J \), is calculated to understand how these transformations distort space. The Jacobian is a determinant formed from the partial derivatives of the new variables \( x \) and \( y \) with respect to \( u \) and \( v \).
The specific Jacobian for our transformation is:
\[ J = \begin{vmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{vmatrix} = \begin{vmatrix} -\frac{1}{2} & \frac{1}{2} \ \frac{1}{2} & \frac{1}{2} \end{vmatrix} = \frac{1}{2} \] This determinant \( \frac{1}{2} \) scales the transformed integral, showing how volumes change under transformation.
In our transformation, where \( u = y-x \) and \( v = x+y \), the Jacobian matrix, denoted \( J \), is calculated to understand how these transformations distort space. The Jacobian is a determinant formed from the partial derivatives of the new variables \( x \) and \( y \) with respect to \( u \) and \( v \).
The specific Jacobian for our transformation is:
\[ J = \begin{vmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{vmatrix} = \begin{vmatrix} -\frac{1}{2} & \frac{1}{2} \ \frac{1}{2} & \frac{1}{2} \end{vmatrix} = \frac{1}{2} \] This determinant \( \frac{1}{2} \) scales the transformed integral, showing how volumes change under transformation.
Completing the Square
The mathematical technique known as completing the square is invaluable for simplification in integrals. In the given problem, the expression \( y^2 - 2xy + x^2 \) closely resembles \((y-x)^2\).
Completing the square means rewriting a quadratic expression in the form \((a-b)^2 + constant\). This form is much more convenient for integration or transformation. By viewing the original expression as \((y-x)^2\), it spurred choosing the substitution \( u = y-x \).
The result was a transformed integral that contained \( e^{u^2} \) as part of its integrand. Thus, starting with this technique helped to not only simplify but also revealed the symmetry necessary for easier evaluation.
Completing the square means rewriting a quadratic expression in the form \((a-b)^2 + constant\). This form is much more convenient for integration or transformation. By viewing the original expression as \((y-x)^2\), it spurred choosing the substitution \( u = y-x \).
The result was a transformed integral that contained \( e^{u^2} \) as part of its integrand. Thus, starting with this technique helped to not only simplify but also revealed the symmetry necessary for easier evaluation.
Limits of Integration
When transforming an integral to new variables, recalculating the limits of integration is crucial. These new limits define the region of integration in terms of the new variables.
Initially, we had bounds \( y=0 \) to \( y=x+2 \) for \( y \), and \( x=-2 \) to \( x=0 \). Using substitutions \( u = y-x \), and noting the original bounds, we can transform these to conditions for \( u \) and \( v \).
For example, transforming the line \( y=x+2 \) results in new bounds. If \( y=0 \) becomes \( u=-x \) and \( y=x+2 \) becomes \( u=2-x \), then \( v \) stretches from \(-2\) to \(2\) owing to the \( x \) limits.
This step ensures that the integration captures the correct region with the new variables, preserving the physical or geometrical interpretation of the original problem.
Initially, we had bounds \( y=0 \) to \( y=x+2 \) for \( y \), and \( x=-2 \) to \( x=0 \). Using substitutions \( u = y-x \), and noting the original bounds, we can transform these to conditions for \( u \) and \( v \).
For example, transforming the line \( y=x+2 \) results in new bounds. If \( y=0 \) becomes \( u=-x \) and \( y=x+2 \) becomes \( u=2-x \), then \( v \) stretches from \(-2\) to \(2\) owing to the \( x \) limits.
This step ensures that the integration captures the correct region with the new variables, preserving the physical or geometrical interpretation of the original problem.
Other exercises in this chapter
Problem 23
Find a vector that gives the direction in which the given function increases most rapidly at the indicated point. Find the maximum rate. $$ f(x, y)=e^{2 x} \sin
View solution Problem 23
Graph the curve \(C\) that is described by \(\mathbf{r}\) and graph \(\mathbf{r}^{\prime}\) at the indicated value of \(\boldsymbol{t}\). $$ \mathbf{r}(t)=2 \ma
View solution Problem 24
Find the volume of the solid bounded by the graphs of the given equations. \(x=2, y=x, y=0, \quad z=x^{2}+y^{2}, \quad z=0\)
View solution Problem 24
Evaluate the surface integral \(\iint_{S} G(x, y, z) d S\). \(G(x, y, z)=\left(1+4 y^{2}+4 z^{2}\right)^{1 / 2} ; S\) that portion of the paraboloid \(x=4-y^{2}
View solution