Change of variables in double integrals

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Change of variables in double integrals

This is a useful technique, particularly when dealing with integrals over regions that are difficult to compute using standard Cartesian coordinates. By transforming to a different coordinate system (like polar, cylindrical, or spherical coordinates) or through a more general transformation, the integral can often be simplified.

Suppose that we want to calculate the area of an ellipse. An ellipse with semiaxes a and b is specified by $(\frac{x}{a})^2 + (\frac{y}{b})^2 = 1$. If r = a = b. then this defines a circle.

We want to calculate $ \iint_{(\frac{x}{a})^2 + (\frac{y}{b})^2 < 1} dxdy$ =[Change of variables, set $\frac{x}{a}=u, \frac{y}{b}=b⇒ du = \frac{1}{a}dx, db = \frac{1}{b}dy$⇒ dxdy = ab·dudv] $ \iint_{u^2 + v^2 < 1} abdudv = ab\iint_{u^2 + v^2 < 1} dudv =$[Area of the unit circle] ab·π.

In this particular case, the change of variables was pretty easy. This is not always the case, suppose that u = 3x -2y, v = x + y. We want to map a region R in the xy-plane to a region S in the uv-plane, understand the relation between dA = dxdy and dA’ = dudv.

Let’s consider ΔA to be a rectangle, ΔA’ is a parallelogram and we know that because the relations between (x, y) and (u, v) are linear and there is an Area scaling factor that does not depend on the choice of the rectangle (Figure 1 and 2).

We will try to understand this with the unit rectangle (Figure 3). Recall that in the context of vector in 2-dimensional space, the determinant $|\begin{smallmatrix}u_1 & v_1\\\ u_2 & v_2\end{smallmatrix}|$ of two vectors $\vec{u}$ = ⟨u₁, u₂⟩, and $\vec{v}$ = (v₁, v₂) represents the signed area of the parallelogram spanned by those vectors.

In our example, A’ = $|\begin{smallmatrix}3 & 1\\ -2 & 1\end{smallmatrix}| = 5$. More importantly, for any arbitrary rectangle, the area scaling factor is always 5, so dA’ = 5dA ⇒ dudv = 5dxdy.

The change of variables formula for double integrals states that if you have a transformation that maps a region R in the xy-plane to a region S in the uv-plane, then the double integral over R can be computed by transforming it into an integral over R with a formula given by $ \iint_R f(x, y)dA = \iint_s f(x(u,v), y(u, v)·factor·dA’ =$[In our example] $\iint_S f(x(u,v), y(u, v))\frac{1}{5}dA'$

In the general case, u = u(x, y), v = (x, y), if we consider small changes Δx and Δy in the xy-plane, the changes Δu and Δv can be approximated by the expressions $Δu ≈ \frac{∂u}{∂x}Δx + \frac{∂u}{∂y}Δy$ = u_xΔx + u_yΔy where $\frac{∂u}{∂x}, \frac{∂u}{∂y}$ denote the partial derivatives of u with respect to x and y respectively (they measure how the function u changes as each variable changes, while holding the other variables constant), and similarly Δv ≈ v_xΔx + v_yΔy, that is, $|\begin{smallmatrix}Δu\\ Δv\end{smallmatrix}| = |\begin{smallmatrix}u_x & u_y\\ v_x & v_y\end{smallmatrix}||\begin{smallmatrix}Δx\\ Δy\end{smallmatrix}|$, so we can conclude the scaling factor is this determinant (Figure 4).

The Jacobian determinant is J = $\frac{∂(u,v)}{∂(x,y)} = |\begin{smallmatrix}u_x & u_y\\ v_x & v_y\end{smallmatrix}|⇒ dudv = |J|dxdy= |\begin{smallmatrix}u_x & u_y\\ v_x & v_y\end{smallmatrix}|dxdy$.

To sum up, the change of variables formulas for double integrals states that if you have a transformation that maps a region R in the xy-plane to a region S in the uv-plane, then the double integral over R can be computed by transforming it into an integral over R with a formula given by $ \iint_R f(x, y)dA = \iint_S f(x(u,v), y(u, v)·\frac{1}{|\begin{smallmatrix}u_x & u_y\\\ v_x & v_y\end{smallmatrix}|}dudv$ where f(x, y) represents the function to be integrated over R, f(x(u, v), y(u, v)) is the same function expressed in terms of the new variables u and v and $\frac{1}{|\begin{smallmatrix}u_x & u_y\\ v_x & v_y\end{smallmatrix}|}$ is the reciprocal of the determinant of the Jacobian matrix of the transformation. It is a factor that adjust the area element for the change in the scale cause by the transformation.

Polar coordinates are a common choice for regions with circular symmetry. When converting from Cartesian coordinates (x, y) to polar coordinates (r, θ), or vice versa, the Jacobian determinant provides the necessary factor to correctly adjust the differential area element.

The transformation equations are x = rcos(θ), y = rsin(θ). The Jacobian matrix of this transformation is $J = \frac{∂(x, y)}{∂(r, θ)} = (\begin{smallmatrix}\frac{∂x}{∂r} & \frac{∂x}{∂θ}\\ \frac{∂y}{∂r} & \frac{∂y}{∂θ}\end{smallmatrix}) = (\begin{smallmatrix}x_r & x_θ\\ y_r & y_θ\end{smallmatrix}) = (\begin{smallmatrix}cos(θ) & -rsin(θ)\\ sin(θ) & rcos(θ)\end{smallmatrix}).$

The Jacobian determinant det(J) = $rcos^2(θ)+rsin^2(θ) = r ⇒ dxdy = rdrdθ$. The Jacobian determinant for this transformation is r, so the area element dA in polar coordinates becomes rdrdθ. Conversely, if you start in polar coordinates and wish to convert an integral back to Cartesian coordinates, you use the inverse of the Jacobian determinant, which for this transformation would simply be 1/r.

• Calculate $\int_{0}^{1} \int_{0}^{1} x^2ydxdy$ by changing to u = x, v = xy

Steps for Changing variables:

1. Identify the new coordinates: u = x, v = xy
2. Determine the Jacobian determinant, $det(J) = \frac{∂(u, v)}{∂(x, y)} = |\begin{smallmatrix}\frac{∂u}{∂x} & \frac{∂u}{∂y}\\ \frac{∂v}{∂x} & \frac{∂v}{∂y}\end{smallmatrix}| = |\begin{smallmatrix}1 & 0\\ y & x\end{smallmatrix}| = x$
3. Setup the integral in the new coordinates. The differential area dA = dudv = x·dxdy.
4. Integrate.

The integrand in terms of u, v: $x^2ydxdy = x^2y\frac{1}{x}dudv = xydudv = vdudv$ ⇒ $\int_{0}^{1}\int_{0}^{1} x^2ydxdy = \int_{0}^{1}\int_{v}^{1} vdudv$ (Figure 1, Figure 2 illustrates another way of looking at the same problem).

The inner integral is $\int_{v}^{1} vdu = v\int_{v}^{1} du = v·u\bigg|_{v}^{1} = v(1-v)$

$\int_{0}^{1}\int_{v}^{1} vdudv = \int_{0}^{1} v(1-v) dv = \int_{0}^{1} v-v^2 dv = \frac{1}{2}v^2-\frac{1}{3}v^3\bigg|_{0}^{1} = \frac{1}{2}-\frac{1}{3} = \frac{3}{6}-\frac{2}{6} =\frac{1}{6}.$

Bibliography

1. NPTEL-NOC IITM, Introduction to Galois Theory.
2. Algebra, Second Edition, by Michael Artin.
3. LibreTexts, Calculus and Calculus 3e (Apex). Abstract and Geometric Algebra, Abstract Algebra: Theory and Applications (Judson).
4. Field and Galois Theory, by Patrick Morandi. Springer.
5. Michael Penn, and MathMajor.
6. Contemporary Abstract Algebra, Joseph, A. Gallian.
7. YouTube’s Andrew Misseldine: Calculus. College Algebra and Abstract Algebra.
8. MIT OpenCourseWare 18.01 Single Variable Calculus, Fall 2007 and 18.02 Multivariable Calculus, Fall 2007.
9. Calculus Early Transcendentals: Differential & Multi-Variable Calculus for Social Sciences.