Chapter 33 — Quiz

Ten questions on the Jacobian determinant and change of variables. Answer each, then expand the solution to check yourself and find the relevant section. Aim to explain each answer, not just match it.

1. The Jacobian determinant $\lvert\det J_T\rvert$ of a transformation $T:(u,v)\to(x,y)$ measures:

  • A) the length scaling along the $u$-axis
  • B) the local area scaling: a cell $du\,dv$ maps to area $\lvert\det J_T\rvert\,du\,dv$
  • C) the angle distortion of the map
  • D) the energy of the transformation
Answer**B.** The Jacobian determinant is the area of the image parallelogram per unit input area, derived in Section 33.6. It is the 2D successor of the single-variable stretch factor $g'(x)$. *Section 33.3, Section 33.6.*

2. For the polar map $x = r\cos\theta$, $y = r\sin\theta$, the Jacobian determinant is:

  • A) $1$ B) $r$ C) $r^2$ D) $\theta$
Answer**B) $r$.** $\det J = \cos\theta(r\cos\theta) - (-r\sin\theta)\sin\theta = r(\cos^2\theta + \sin^2\theta) = r$, giving $dA = r\,dr\,d\theta$. The factor grows with $r$ because a wedge of fixed angular width spans more arc the farther out you go. *Section 33.3.*

3. The spherical volume element (with $\phi$ the angle from the positive $z$-axis) is:

  • A) $r\,dr\,d\theta\,dz$
  • B) $\rho^2\,d\rho\,d\phi\,d\theta$
  • C) $\rho^2\sin\phi\,d\rho\,d\phi\,d\theta$
  • D) $\rho\sin\phi\,d\rho\,d\phi\,d\theta$
Answer**C.** $\det J = \rho^2\sin\phi$. The $\rho^2$ says volume grows like the square of the radius; the $\sin\phi$ pinches wedges thin near the poles and fattest at the equator. *Section 33.7.*

4. In the 2D change-of-variables formula, the area element transforms as $dx\,dy =$

  • A) $du\,dv$ B) $\det J\,du\,dv$ C) $\lvert\det J\rvert\,du\,dv$ D) $J\,du\,dv$
Answer**C) $\lvert\det J\rvert\,du\,dv$.** The absolute value is the only genuinely new wrinkle over single-variable substitution: the *sign* of the determinant records orientation, which area discards. *Section 33.4.*

5. If $T:(u,v)\to(x,y)$ is invertible, the Jacobian determinant of $T^{-1}$ equals:

  • A) $\det J_T$ B) $1/\det J_T$ C) $-\det J_T$ D) undefined
Answer**B) $1/\det J_T$.** Since $J_{T^{-1}}J_T = I$ (chain rule) and the determinant is multiplicative, the two determinants are reciprocals — the multivariable echo of $\frac{dx}{du}\cdot\frac{du}{dx}=1$. *Section 33.8.*

6. The Inverse Function Theorem guarantees a local smooth inverse of $T$ at a point $p$ provided:

  • A) $\det J_T(p) = 0$
  • B) $\det J_T(p) \neq 0$
  • C) $\det J_T(p) = 1$
  • D) no condition is needed
Answer**B) $\det J_T(p)\neq 0$.** A nonzero Jacobian determinant is a *certificate of local invertibility*, the higher-dimensional version of "$f'(a)\neq 0 \Rightarrow f$ locally invertible." Where $\det J_T = 0$, the map collapses dimensions. *Section 33.8.*

7. To find the area of the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2}\le 1$, use $x = au$, $y = bv$. The area is:

  • A) $\pi a^2$ B) $\pi b^2$ C) $\pi ab$ D) $2\pi(a+b)$
Answer**C) $\pi ab$.** The stretch sends the unit disk (area $\pi$) to the ellipse, and $\det J = ab$, so area $= ab\cdot\pi = \pi ab$. *Section 33.5.*

8. Worked Example 4 evaluates $\iint_{\mathbb{R}^2} e^{-(x^2+y^2)}\,dA$ by switching to polar. The polar Jacobian factor of $r$ matters because:

  • A) it cancels the $e^{-r^2}$ in the integrand
  • B) it makes the radial integral $\int_0^\infty e^{-r^2}r\,dr$ elementary (via $s=r^2$), whereas $\int e^{-x^2}dx$ has no elementary antiderivative
  • C) it is required to keep the angle bounded
  • D) it changes the value of the integral
Answer**B.** The factor of $r$ from the Jacobian is exactly what makes $\int_0^\infty e^{-r^2}r\,dr = \tfrac12$ solvable; the whole 2D integral is then $\pi$, from which $\int_{-\infty}^\infty e^{-x^2}dx = \sqrt{\pi}$ follows. *Section 33.4.*

9. A pair of random variables $(X,Y)$ with joint density $p_{XY}$ is mapped to $(U,V) = T(X,Y)$. The new density is:

  • A) $p_{UV}(u,v) = p_{XY}(x(u,v),y(u,v))$
  • B) $p_{UV}(u,v) = p_{XY}(x(u,v),y(u,v))\,\lvert\det J_{T^{-1}}\rvert$
  • C) $p_{UV}(u,v) = p_{XY}(x(u,v),y(u,v)) + \lvert\det J_{T^{-1}}\rvert$
  • D) $p_{UV}(u,v) = \lvert\det J_T\rvert$
Answer**B.** A density is "probability per unit area," so when area rescales the density must rescale inversely. Here $T^{-1}$ expresses the old variables in terms of the new, and its Jacobian factor keeps the total probability equal to $1$. *Section 33.9.*

10. Modern normalizing flows (Real NVP, Glow) build a complex distribution by pushing a Gaussian through invertible maps $T$, computing $p(\mathbf{x}) = p_{\text{base}}(T^{-1}(\mathbf{x}))\,\lvert\det J_{T^{-1}}\rvert$. The central engineering goal is to design $T$ so that:

  • A) $\det J$ is always exactly $1$
  • B) $\det J$ is cheap to compute (often by making $J$ triangular, so $\det J$ is the product of the diagonal)
  • C) $T$ is linear
  • D) $T$ has no inverse
Answer**B.** Expressiveness must be balanced against a tractable $\det J$. Triangular Jacobians make the determinant a product of diagonal entries, so the change-of-variables formula — almost verbatim the loss function — stays cheap. *Section 33.9.*

Scoring Guide

Score Interpretation
9–10 Excellent. You can compute Jacobians, apply the formula in 2D/3D, and explain the density and orientation subtleties. Move on to Chapter 34.
7–8 Solid. Re-read the section flagged on any miss — most likely the spherical factor (Q3) or the density direction (Q9).
5–6 Partial. Re-derive the polar, cylindrical, and spherical Jacobians by hand (Section 33.3, Section 33.7) before continuing.
0–4 Review the chapter. Focus on Section 33.3 (what the Jacobian is), Section 33.4 (the formula), and Section 33.6 (why it is the area factor).