Change of Order of Integration & Change of Variables via Jacobians

Explanation

Many practical definite integrals (such as ∫ (sin y / y) dy or ∫ e^{-x²} dx) have no elementary antiderivative. By changing the order of integration, the non-integrable variable becomes the outer parameter, yielding an inner integrand of 1 that integrates trivially. When transforming coordinate geometries (e.g., from skewed parallelograms to unit squares), the Jacobian determinant measures the local surface area distortion factor |dA_{xy} / dA_{uv}|.

Key points

• Procedure: (1) Extract existing boundary equations from integral limits, (2) Sketch the exact 2D bounded region $R$, (3) Switch perspective from vertical strips to horizontal strips (or vice-versa), (4) Determine new inner variable limits as functions and outer limits as constants.
• Change of Variables in Double Integrals: ${\iint }_{R}f(x,y)dxdy={\iint }_{R^{\prime }}f(x(u,v),y(u,v))|J|dudv$.
• The Jacobian Determinant: $J=\frac{\partial (x,y)}{\partial (u,v)}=\left|\begin{array}{cc}\frac{\partial x}{\partial u}& \frac{\partial x}{\partial v}\\ \frac{\partial y}{\partial u}& \frac{\partial y}{\partial v}\end{array}\right|=\frac{\partial x}{\partial u}\frac{\partial y}{\partial v}-\frac{\partial x}{\partial v}\frac{\partial y}{\partial u}$.
• Inverse Jacobian Property: $J\cdot J^{\prime }=1$, meaning $\frac{\partial (x,y)}{\partial (u,v)}=\frac{1}{\frac{\partial (u,v)}{\partial (x,y)}}$.
• Polar Coordinate Transformation: $x=r\cos \theta$, $y=r\sin \theta ⟹$ Jacobian $J=r⟹dxdy=rdrd\theta$.

Code examples

// Change of Order of Integration: Classic University Exam Proof
// Evaluate: I = ∫_0^a ∫_{x/a}^{√(x/a)} (x^2 + y^2) dy dx
// Step 1: Identify existing boundary curves:
//    Inner limits: y = x/a (straight line x = ay) to y = √(x/a) (parabola x = a y^2)
//    Outer limits: x = 0 to x = a
// Step 2: Points of intersection:
//    ay = ay^2 => ay(1 - y) = 0 => y = 0 and y = 1.
//    At y = 0, x = 0. At y = 1, x = a.
// Step 3: Draw horizontal strips across region R:
//    For a fixed y ∈ [0, 1]:
//    Left boundary curve (enters):  x = a y^2
//    Right boundary curve (exits): x = a y
// Step 4: Rewritten integral with changed order:
//    I = ∫_0^1 [ ∫_{a y^2}^{a y} (x^2 + y^2) dx ] dy
//      = ∫_0^1 [ (x^3 / 3) + y^2 x ]_{a y^2}^{a y} dy
//      = ∫_0^1 [ (a^3 y^3 / 3 + a y^3) - (a^3 y^6 / 3 + a y^4) ] dy
// Step 5: Integrate with respect to y:
//    = [ (a^3/3 + a) (y^4 / 4) - (a^3 / 21) y^7 - (a / 5) y^5 ]_0^1
//    = (a^3 / 12) + (a / 4) - (a^3 / 21) - (a / 5)
//    = a^3 (1/12 - 1/21) + a (1/4 - 1/5) = (3 a^3 / 84) + (a / 20) = (a^3 / 28) + (a / 20)

Exam questions

• [8 Marks] Change the order of integration and evaluate ∫_0^1 ∫_x^1 [x / (x² + y²)] dy dx.
• [8 Marks] Change the order of integration of ∫0^4a ∫{x²/4a}^{2√(ax)} dy dx and evaluate.
• [7 Marks] Transform to polar coordinates and evaluate ∬_R √(a² - x² - y²) dx dy over the positive quadrant of the circle x² + y² ≤ a².
• [7 Marks] Find the Jacobian ∂(u, v)/∂(x, y) if u = (x² - y²) / 2 and v = xy.