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Calculus / Chinese

Substitution in integrals

Charles队长 · Bilibili · 0:24

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The explanation, unpacked.

Reviewed learning material · Video analysis · English
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The video illustrates the substitution method for double integrals using polar coordinates. It contrasts Cartesian and parameter planes, mapping a circular region to a rectangular one. The segment derives the transformation equations, calculates the Jacobian determinant as 'r', and presents the final change-of-variable formula.

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Chapters

0:00Coordinate Systems and Region Mapping0:14Transformation Equations and Jacobian0:20Double Integral Substitution Formula

Learning script

Generated from the video's visuals and explanation; not verbatim speech.

The screen splits into two views: a standard Cartesian coordinate system (xOyxOy) on the left and a parameter plane defined by angle θ\theta and radius rr on the right. An animation draws a closed disk centered at the origin with radius 1 (labeled r=1r=1) on the left grid. Simultaneously, a red rectangle appears on the right, spanning horizontally from 0 to 1 and vertically from 0 to 2π2\pi. This visualizes how a complex curved boundary transforms into a simple rectangular domain in the new variables.

Above the graphs, the polar coordinate transformation equations appear within braces: x=rcos⁡θx = r \cos \theta and y=rsin⁡θy = r \sin \theta. Immediately following this, the Jacobian determinant is displayed: J=∣cos⁡θ−rsin⁡θsin⁡θrcos⁡θ∣J = \begin{vmatrix} \cos \theta & -r \sin \theta \\ \sin \theta & r \cos \theta \end{vmatrix}. The calculation simplifies directly to =r= r, demonstrating the necessary scaling factor required when converting area elements between these systems.

A change of variables updates the integrand, the region and the area factor |J|=r together. The integral remains two-dimensional; the description may simply become easier to use.

Knowledge cards

01

Polar Coordinate Transformation

Defines the relationship between Cartesian coordinates and polar parameters. This substitution is essential for integrating functions over regions exhibiting radial symmetry or bounded by circles.

{x=rcos⁡θy=rsin⁡θ\begin{cases} x = r \cos \theta \\ y = r \sin \theta \end{cases}
02

Role of the Jacobian Determinant

Represents the local magnification factor of area during variable substitution. In polar coordinates, it accounts for the fact that arc length increases with distance from the origin, necessitating multiplication by rr.

J=∣∂(x,y)∂(r,θ)∣=rJ = \left| \frac{\partial(x,y)}{\partial(r,\theta)} \right| = r
03

Change of Variables Rule

General principle stating that an integral over a domain DD equals the integral over transformed domain D′D' of the composed function multiplied by the absolute value of the Jacobian.

∬Df(x,y)dxdy=∬D′f(ϕ1(u,v),ϕ2(u,v))∣J∣dudv\iint_D f(x,y) dx dy = \iint_{D'} f(\phi_1(u,v), \phi_2(u,v)) |J| du dv
04

Domain Correspondence

Parameterize the unit disk by 0≤r≤10\le r\le 1 and 0≤θ<2π0\le θ<2π. The origin and angular seam prevent a global one-to-one smooth map, but these zero-area exceptions do not affect the usual integral. A rectangular parameter domain does not itself make the integrand separable.

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  • Determinants ApplicationAt 0:17
    Why this connection?

    The polar substitution computes the Jacobian determinant r and uses its absolute value as the local area scale; r≥0r\ge 0 makes the factor r. The unit-disk parameterization has zero-area exceptions at the origin and angular seam. A rectangular parameter domain alone does not make the integrand separable.