Multivariable-Calculus · Unit 16 · Video 2 · Interactive Practice

The Jacobian: Why Every Polar Integral Carries an r

IKey Formulas

FormulaNameWhat it says
J=∂(u,v)∂(x,y)=∣uxuyvxvy∣J = \dfrac{\partial(u,v)}{\partial(x,y)} = \begin{vmatrix} u_x & u_y \\ v_x & v_y \end{vmatrix}JacobianDeterminant of the matrix of partial derivatives
du dv=∣J∣ dx dydu\,dv = |J|\,dx\,dyLocal area scale factorAbsolute value: area is never negative
∂(u,v)∂(x,y)⋅∂(x,y)∂(u,v)=1\dfrac{\partial(u,v)}{\partial(x,y)} \cdot \dfrac{\partial(x,y)}{\partial(u,v)} = 1Reciprocal JacobiansCompute whichever is easier
∂(x,y)∂(r,θ)=r ⇒ dx dy=r dr dθ\dfrac{\partial(x,y)}{\partial(r,\theta)} = r \ \Rightarrow\ dx\,dy = r\,dr\,d\thetaPolar coordinatesThe rr in every polar integral

Key Insight: near any point a curved substitution is approximately linear, with the matrix of partials in place of a constant matrix, so its area scale factor ∣J∣|J| changes from point to point.

IIEqual Cells, Unequal Images

Every cell of the rθr\theta-grid has the same Δr Δθ\Delta r\,\Delta\theta, but its image in the xyxy-plane has area close to r Δr Δθr\,\Delta r\,\Delta\theta.

IIIShrinking to the Jacobian

As the step hh shrinks, the curved image of an h×hh \times h cell becomes its tangent parallelogram, and both areas approach r h2r\,h^2.

IVQuiz Questions

Problem 1 · The Scale Factor at a Point

Given: u=x2+y, v=x−y2u = x^2 + y,\ v = x - y^2 — find kk in du dv≈k dx dydu\,dv \approx k\,dx\,dy near the point (x,y)=(1,2)(x, y) = (1, 2).

✅ Correct! J=−4xy−1=−9J = -4xy - 1 = -9 at (1,2)(1,2), and the area factor is ∣J∣=9|J| = 9.
❌ Check the sign. J=−9J = -9 is right, but du dv=∣J∣ dx dydu\,dv = |J|\,dx\,dy: an area factor is never negative.
❌ Not quite. The determinant subtracts the off-diagonal product: uxvy−uyvx=(2)(−4)−(1)(1)=−9u_x v_y - u_y v_x = (2)(-4) - (1)(1) = -9, not −8+1-8 + 1.
❌ Wrong direction. ∂(u,v)/∂(x,y)\partial(u,v)/\partial(x,y) gives du dvdu\,dv per dx dydx\,dy; 19\tfrac19 is the factor in dx dy=19 du dvdx\,dy = \tfrac19\,du\,dv.
Show solution

The partials are ux=2xu_x = 2x, uy=1u_y = 1, vx=1v_x = 1, vy=−2yv_y = -2y, so

J=∂(u,v)∂(x,y)=∣2x11−2y∣=−4xy−1.J = \frac{\partial(u,v)}{\partial(x,y)} = \begin{vmatrix} 2x & 1 \\ 1 & -2y \end{vmatrix} = -4xy - 1.

At (1,2)(1, 2): J=−8−1=−9J = -8 - 1 = -9, so du dv≈∣−9∣ dx dy=9 dx dydu\,dv \approx |{-9}|\,dx\,dy = 9\,dx\,dy.

Problem 2 · Which Jacobian to Use

Given: x=u2−v2, y=2uvx = u^2 - v^2,\ y = 2uv — find the expression to replace dx dydx\,dy with when an integral is switched to u,vu, v.

✅ Correct! Here x,yx, y are given as functions of u,vu, v, so dx dy=∣∂(x,y)∂(u,v)∣du dv=4(u2+v2) du dvdx\,dy = \left|\frac{\partial(x,y)}{\partial(u,v)}\right| du\,dv = 4(u^2 + v^2)\,du\,dv.
❌ Wrong direction. With x,yx, y written in terms of u,vu, v, the Jacobian you compute directly is ∂(x,y)/∂(u,v)\partial(x,y)/\partial(u,v), and it multiplies du dvdu\,dv.
❌ Check the determinant. xuyv−xvyu=(2u)(2u)−(−2v)(2v)=4u2+4v2x_u y_v - x_v y_u = (2u)(2u) - (-2v)(2v) = 4u^2 + 4v^2: subtracting a negative adds.
❌ Not quite. Each partial carries a factor 22, so the determinant carries 2⋅2=42 \cdot 2 = 4.
Show solution
∂(x,y)∂(u,v)=∣xuxvyuyv∣=∣2u−2v2v2u∣=4u2+4v2.\frac{\partial(x,y)}{\partial(u,v)} = \begin{vmatrix} x_u & x_v \\ y_u & y_v \end{vmatrix} = \begin{vmatrix} 2u & -2v \\ 2v & 2u \end{vmatrix} = 4u^2 + 4v^2.

This is never negative, so dx dy=4(u2+v2) du dvdx\,dy = 4(u^2 + v^2)\,du\,dv.

Problem 3 · Using the Reciprocal

Given: u=xy, v=y/xu = xy,\ v = y/x on x,y>0x, y > 0 — find dx dydx\,dy in terms of uu, vv and du dvdu\,dv.

✅ Correct! ∂(u,v)/∂(x,y)=2y/x=2v\partial(u,v)/\partial(x,y) = 2y/x = 2v, and its reciprocal gives dx dy=12v du dvdx\,dy = \frac{1}{2v}\,du\,dv.
❌ Wrong direction. 2v2v is ∂(u,v)/∂(x,y)\partial(u,v)/\partial(x,y), the factor in du dv=2v dx dydu\,dv = 2v\,dx\,dy; solve for dx dydx\,dy.
❌ Not quite. The Jacobian is 2y/x2y/x, and y/xy/x is vv, not u=xyu = xy.
❌ Not quite. The two terms of the determinant add: y⋅1x−x⋅(−yx2)=2yxy \cdot \frac1x - x \cdot \left(-\frac{y}{x^2}\right) = \frac{2y}{x}, which depends on the point.
Show solution

It is easier to differentiate u,vu, v with respect to x,yx, y than to solve for x,yx, y:

∂(u,v)∂(x,y)=∣yx−y/x21/x∣=yx+yx=2yx=2v.\frac{\partial(u,v)}{\partial(x,y)} = \begin{vmatrix} y & x \\ -y/x^2 & 1/x \end{vmatrix} = \frac{y}{x} + \frac{y}{x} = \frac{2y}{x} = 2v.

So du dv=2v dx dydu\,dv = 2v\,dx\,dy (v>0v > 0 here), and by the reciprocal relation dx dy=12v du dvdx\,dy = \dfrac{1}{2v}\,du\,dv.

Problem 4 · Elliptical Polar Coordinates

Given: x=arcos⁡θ, y=brsin⁡θx = a r\cos\theta,\ y = b r\sin\theta with a,b>0a, b > 0 — find the Jacobian ∂(x,y)/∂(r,θ)\partial(x,y)/\partial(r,\theta).

✅ Correct! The determinant is abr(cos⁡2θ+sin⁡2θ)=abrabr(\cos^2\theta + \sin^2\theta) = abr, so the ellipse's area is ∫02π ⁣∫01abr dr dθ=πab\int_0^{2\pi}\!\int_0^1 abr\,dr\,d\theta = \pi ab.
❌ Not quite. That is the ordinary polar Jacobian; the stretches by aa and bb multiply it by abab.
❌ Not quite. The θ\theta-column carries a factor rr: xθ=−arsin⁡θx_\theta = -ar\sin\theta and yθ=brcos⁡θy_\theta = br\cos\theta.
❌ Check the sign. xryθ−xθyr=abrcos⁡2θ−(−abrsin⁡2θ)x_r y_\theta - x_\theta y_r = abr\cos^2\theta - (-abr\sin^2\theta): the two squares add to 11.
Show solution
∂(x,y)∂(r,θ)=∣acos⁡θ−arsin⁡θbsin⁡θbrcos⁡θ∣=abrcos⁡2θ+abrsin⁡2θ=abr.\frac{\partial(x,y)}{\partial(r,\theta)} = \begin{vmatrix} a\cos\theta & -ar\sin\theta \\ b\sin\theta & br\cos\theta \end{vmatrix} = abr\cos^2\theta + abr\sin^2\theta = abr.

So dx dy=abr dr dθdx\,dy = abr\,dr\,d\theta. With 0≤r≤10 \le r \le 1, 0≤θ≤2π0 \le \theta \le 2\pi this sweeps the ellipse (x/a)2+(y/b)2≤1(x/a)^2 + (y/b)^2 \le 1, of area ∫02π ⁣∫01abr dr dθ=πab\int_0^{2\pi}\!\int_0^1 abr\,dr\,d\theta = \pi ab.

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