Jacobian & Coordinate Transformations - Calculus 3

1. A student claims that for any transformation $T: (x,y) \mapsto (u,v)$, if the Jacobian determinant $\frac{\partial(u,v)}{\partial(x,y)} = 0$ at a point, then the transformation cannot be used to evaluate double integrals in a neighborhood of that point. Which statement best evaluates this claim?
   Answer: The claim is correct because zero Jacobian means the transformation is not locally invertible When you encounter questions about Jacobian determinants and coordinate transformations, you're dealing with the fundamental requirement for valid change of variables in multiple integrals. The key insight is that transformations must be locally invertible to preserve the integral's meaning. The Jacobian determinant $\frac{\partial(u,v)}{\partial(x,y)}$ measures how the transformation stretches or compresses area elements. When this determinant equals zero, the transformation collapses a 2D region into a lower-dimensional object (like a curve or point). This means the transformation isn't locally invertible—you can't uniquely determine the original $(x,y)$ coordinates from the transformed $(u,v)$ coordinates in that neighborhood. Without local invertibility, the change of variables formula breaks down because you need a one-to-one correspondence between regions to properly transform the integral. Answer D correctly identifies this fundamental requirement. Answer A misses the point entirely—the Jacobian's sign (which determines orientation) is irrelevant when the determinant is zero. Answer B suggests using limits, but this doesn't resolve the underlying problem that the transformation fails to be invertible. Answer C incorrectly distinguishes between linear and nonlinear transformations; both require non-zero Jacobian for local invertibility. Study tip: Remember that for any coordinate transformation in multivariable calculus, non-zero Jacobian determinant is essential for local invertibility. Zero Jacobian is a red flag that signals the transformation won't work for integration in that region—this applies universally, regardless of whether the transformation is linear or nonlinear.
2. Let $T$ be the transformation defined by $u = 2x + y$ and $v = x - 3y$. If the Jacobian determinant $\frac{\partial(u,v)}{\partial(x,y)} = -7$, what is the Jacobian determinant $\frac{\partial(x,y)}{\partial(u,v)}$ for the inverse transformation?
   Answer: $-\frac{1}{7}$ For a transformation and its inverse, the Jacobian determinants are reciprocals: $\frac{\partial(x,y)}{\partial(u,v)} = \frac{1}{\frac{\partial(u,v)}{\partial(x,y)}} = \frac{1}{-7} = -\frac{1}{7}$. This relationship holds because the Jacobian matrix of the inverse transformation is the inverse of the original Jacobian matrix, and the determinant of an inverse matrix is the reciprocal of the determinant of the original matrix.
3. Consider the transformation $x = u + v$, $y = u - v$. A student evaluates $\iint_R (x^2 + y^2) \, dx \, dy$ by substituting $x^2 + y^2 = (u+v)^2 + (u-v)^2 = 2u^2 + 2v^2$ and claims the integral becomes $\iint_S 2(u^2 + v^2) \, du \, dv$. What error, if any, did the student make?
   Answer: Error in Jacobian; the integral should be $\iint_S 2(u^2 + v^2) \cdot 2 \, du \, dv$ The algebra is correct: $(u+v)^2 + (u-v)^2 = u^2 + 2uv + v^2 + u^2 - 2uv + v^2 = 2u^2 + 2v^2$. However, the student forgot to include the Jacobian determinant. For $x = u + v$, $y = u - v$, we have $\frac{\partial(x,y)}{\partial(u,v)} = \begin{vmatrix} 1 & 1 \\ 1 & -1 \end{vmatrix} = -1 - 1 = -2$, so $|J| = 2$. Therefore, $dx \, dy = 2 \, du \, dv$, and the correct integral is $\iint_S 2(u^2 + v^2) \cdot 2 \, du \, dv$.
4. Consider the transformation $u = e^x \cos y$ and $v = e^x \sin y$. If we want to evaluate $\iint_S g(u,v) \, du \, dv$ by transforming to the $xy$-plane, which expression correctly represents this integral?
   Answer: $\iint_R g(e^x \cos y, e^x \sin y) \cdot e^x \, dx \, dy$ To transform the integral, we need the Jacobian $\frac{\partial(u,v)}{\partial(x,y)}$. Computing: $\frac{\partial u}{\partial x} = e^x \cos y$, $\frac{\partial u}{\partial y} = -e^x \sin y$, $\frac{\partial v}{\partial x} = e^x \sin y$, $\frac{\partial v}{\partial y} = e^x \cos y$. The Jacobian determinant is $|J| = (e^x \cos y)(e^x \cos y) - (-e^x \sin y)(e^x \sin y) = e^{2x}\cos^2 y + e^{2x}\sin^2 y = e^{2x}(\cos^2 y + \sin^2 y) = e^{2x}$. Wait, this gives us choice (c). Let me recalculate more carefully: $|J| = e^x \cos y \cdot e^x \cos y - (-e^x \sin y)(e^x \sin y) = e^{2x}\cos^2 y + e^{2x}\sin^2 y = e^{2x}$. Actually, the determinant calculation was wrong in the original explanation. The correct answer should be (c), not (a).
5. Consider the transformation $u = x + 2y$, $v = 3x + ky$ where $k$ is a constant. For what value of $k$ will this transformation be non-invertible (i.e., the Jacobian determinant equals zero)?
   Answer: $k = 6$ The Jacobian matrix is $J = \begin{pmatrix} 1 & 2 \\ 3 & k \end{pmatrix}$. The determinant is $|J| = 1 \cdot k - 2 \cdot 3 = k - 6$. For the transformation to be non-invertible, we need $|J| = 0$, so $k - 6 = 0$, which gives $k = 6$. When $k = 6$, the transformation equations become linearly dependent.
6. The transformation $u = x^2 - y^2$ and $v = 2xy$ maps a region $R$ in the $xy$-plane to a region $S$ in the $uv$-plane. At the point $(x,y) = (2,1)$, what is the absolute value of the Jacobian determinant $|\frac{\partial(u,v)}{\partial(x,y)}|$?
   Answer: $20$ We compute the partial derivatives: $\frac{\partial u}{\partial x} = 2x$, $\frac{\partial u}{\partial y} = -2y$, $\frac{\partial v}{\partial x} = 2y$, $\frac{\partial v}{\partial y} = 2x$. The Jacobian determinant is $\frac{\partial(u,v)}{\partial(x,y)} = (2x)(2x) - (-2y)(2y) = 4x^2 + 4y^2 = 4(x^2 + y^2)$. At $(2,1)$: $|J| = 4(2^2 + 1^2) = 4(4 + 1) = 4(5) = 20$.
7. For the elliptical transformation $u = \frac{x}{a}$ and $v = \frac{y}{b}$ where $a, b > 0$, suppose we transform the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ to the unit circle $u^2 + v^2 = 1$. If the area of the ellipse is $\pi ab$, what should be the area of the unit circle calculated using the Jacobian relationship?
   Answer: $\pi$ When you encounter coordinate transformations in multivariable calculus, the key insight is understanding how areas scale through the Jacobian determinant. The Jacobian tells us exactly how much a transformation stretches or compresses regions. For the transformation $u = \frac{x}{a}$ and $v = \frac{y}{b}$, let's find the Jacobian. We need $\frac{\partial(x,y)}{\partial(u,v)}$ since we're going from $(u,v)$ back to $(x,y)$. From the given equations, $x = au$ and $y = bv$, so: $J = \begin{vmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{vmatrix} = \begin{vmatrix} a & 0 \\ 0 & b \end{vmatrix} = ab$ The area transformation formula states: Area in $(x,y)$ = Area in $(u,v)$ × $|J|$. Since the ellipse has area $\pi ab$ and $|J| = ab$, the unit circle's area must be $\frac{\pi ab}{ab} = \pi$. This confirms answer (B). Choice (A) $\pi ab$ incorrectly assumes the areas are equal, ignoring the transformation's scaling effect. Choice (C) $\frac{\pi}{ab}$ results from using the wrong Jacobian direction—using $\frac{\partial(u,v)}{\partial(x,y)} = \frac{1}{ab}$ instead of its inverse. Choice (D) $\frac{\pi ab}{a+b}$ appears to confuse the Jacobian determinant with some arithmetic combination of $a$ and $b$. Study tip: Always remember that Jacobian determinants tell you how areas scale. If you know one area and the Jacobian, you can find the corresponding area in the other coordinate system.
8. A transformation $T$ maps the unit square $[0,1] \times [0,1]$ in the $xy$-plane to a parallelogram in the $uv$-plane. If the area of the parallelogram is 15 and the transformation is linear with constant Jacobian determinant, what is $|\frac{\partial(u,v)}{\partial(x,y)}|$?
   Answer: $15$ For a linear transformation with constant Jacobian determinant, the area scaling factor is exactly $|\frac{\partial(u,v)}{\partial(x,y)}|$. Since the unit square has area 1 and is transformed to a parallelogram with area 15, we have: Area of image = $|\frac{\partial(u,v)}{\partial(x,y)}| \times$ Area of original region. Therefore, $15 = |\frac{\partial(u,v)}{\partial(x,y)}| \times 1$, so $|\frac{\partial(u,v)}{\partial(x,y)}| = 15$.
9. Consider the transformation to cylindrical coordinates: $x = r\cos\theta$, $y = r\sin\theta$, $z = z$. When evaluating the triple integral $\iiint_E f(x,y,z) \, dV$ using this transformation, which of the following correctly represents the volume element $dV$ in cylindrical coordinates?
   Answer: $dV = r \, dr \, d\theta \, dz$ The Jacobian determinant for cylindrical coordinates is $|J| = r$. This comes from the $3 \times 3$ Jacobian matrix of the transformation, which yields $\frac{\partial(x,y,z)}{\partial(r,\theta,z)} = r$. Therefore, $dV = dx \, dy \, dz = r \, dr \, d\theta \, dz$. Choice (a) omits the Jacobian, (b) uses $r^2$ which would be for spherical coordinates, and (d) has the reciprocal.

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