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📝 Jacobian determinant in two variables (15 MCQs)

📖 From Calculus • 15. Multiple Integrals Calculus • 15 questions available

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What is Jacobian determinant in two variables?

Definition:
The Jacobian determinant for $x=x(u,v), y=y(u,v)$ is $J = \frac{\partial(x,y)}{\partial(u,v)} = \begin{vmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{vmatrix}$ .

Example:
For polar coords, $J = \begin{vmatrix} \cos\theta & -r\sin\theta \\ \sin\theta & r\cos\theta \end{vmatrix} = r$ .

Reason:
The Jacobian measures the local scaling factor of the transformation, crucial for correcting the area element in the new coordinates.

Easy

Medium

Hard

📝 All Jacobian determinant in two variables MCQs

Q1. Let $u=x+y$ and $v=x-y$ . Which statement best explains why the Jacobian is useful when transforming a double integral?

A.It changes the function being integrated but leaves area unchanged

B.It measures how a small area in the

uv

-plane is locally stretched or compressed in the

xy

-plane ✅

C.It guarantees that every transformation maps a rectangle to a rectangle

D.It removes the need to transform the limits of integration

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The Jacobian describes the local area-scaling factor produced by a transformation. For this mapping, small regions in the $uv$ -plane generally correspond to differently scaled regions in the $xy$ -plane. Therefore, the Jacobian must be included when changing variables in a double integral. It does not eliminate the need to transform the region.

Q2. For $u=x+y$ and $v=x-y$ , what is $\left|\frac{\partial(x,y)}{\partial(u,v)}\right|$ ?

A.-2

\frac{1}{2}

✅

C.-1

-\frac{1}{2}

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: Solving the equations gives $x=\frac{u+v}{2}$ and $y=\frac{u-v}{2}$ . Differentiating these expressions gives a determinant of $-\frac{1}{2}$ . Because area scaling uses the absolute value of the determinant, the required factor is $\frac{1}{2}$ .

Q3. A transformation satisfies $\frac{\partial(u,v)}{\partial(x,y)}=5$ at a point. Assuming the transformation is locally invertible there, what is the corresponding area-scaling factor from the $uv$ -plane back to the $xy$ -plane?

A.-5

-5

\frac{1}{5}

✅

D.-25

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: For a locally invertible transformation, the Jacobian of the inverse transformation is the reciprocal of the original Jacobian, provided the original determinant is nonzero. Since the given determinant is $5$ , the inverse transformation has determinant $\frac{1}{5}$ . The absolute value gives the same positive area-scaling factor.

Q4. A student claims that if $\frac{\partial(u,v)}{\partial(x,y)}=0$ at one point, the transformation must map the entire plane onto a single curve. What is the best evaluation of this reasoning?

A.Correct, because a zero Jacobian always means the whole transformation is one-dimensional

B.Correct, because the transformation cannot be used anywhere

C.Incorrect, because a zero Jacobian at one point only indicates local degeneracy there; behavior elsewhere may be different ✅

D.Incorrect, because the Jacobian is unrelated to dimensionality

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: A zero Jacobian at a particular point indicates that the transformation loses local area information there. It does not imply that the entire mapping collapses globally. Other points may have nonzero Jacobians and remain locally invertible. The student's conclusion improperly extends a local condition to the whole domain.

Q5. Consider $u=x^2-y^2$ and $v=2xy$ . A student computes the Jacobian as $4x^2+4y^2$ and concludes that the transformation is locally invertible everywhere except at the origin. What should be concluded?

A.The student is correct because the Jacobian is positive away from the origin

B.The student is incorrect because the Jacobian has the opposite sign, but local invertibility still holds away from the origin ✅

C.The student is incorrect because the Jacobian is always zero

D.The student is correct only when

x=y

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: Differentiating gives $u_x=2x,\ u_y=-2y,\ v_x=2y,\ v_y=2x$ . The determinant is $4x^2+4y^2$ , which is positive except at $(0,0)$ . Thus the computed expression and conclusion are correct. The determinant does not need to be negative; a nonzero positive determinant also establishes local invertibility.

Q6. Suppose a region in the $xy$ -plane is bounded by $x+y=1,\ x+y=3,\ x-y=0,\ x-y=2$ . A transformation uses $u=x+y$ and $v=x-y$ . Which transformed region is most appropriate?

A.A triangle with vertices

(0,0),(1,0),(0,1)

B.A rectangle described by

1\le u\le3,\ 0\le v\le2

✅

C.A circle described by

u^2+v^2\le4

D.A rectangle described by

0\le u\le1,\ 2\le v\le3

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Each pair of boundary lines becomes a constant-coordinate line under the transformation. The lines $x+y=1$ and $x+y=3$ become $u=1$ and $u=3$ , while $x-y=0$ and $x-y=2$ become $v=0$ and $v=2$ . Therefore, the transformed region is the stated rectangle.

Q7. A rectangular region in the $uv$ -plane has area $12$ . At every point of its image, the absolute Jacobian $\left|\frac{\partial(x,y)}{\partial(u,v)}\right|$ equals $3$ . What is the area of the corresponding region in the $xy$ -plane?

A.-4

B.-9

C.-36 ✅

D.-144

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: The absolute Jacobian gives the local factor by which an area element in the $uv$ -plane changes when mapped into the $xy$ -plane. Since that factor is constantly $3$ , the total area is multiplied by $3$ . Thus an original area of $12$ becomes $12\times3=36$ .

Q8. A designer wants coordinates whose constant-coordinate curves align with the boundaries $xy=2$ and $x/y=3$ in the first quadrant. Which choice is most natural for simplifying the region?

u=x+y,\ v=x-y

u=xy,\ v=x/y

✅

u=x^2+y^2,\ v=x-y

u=x^2-y^2,\ v=xy

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: When boundaries are given directly by expressions such as $xy=\text{constant}$ and $x/y=\text{constant}$ , choosing those expressions as new variables converts the boundaries into coordinate lines. This is a modeling strategy: choose variables that match the geometry of the region rather than selecting variables arbitrarily.

Q9. For $u=x+y$ and $v=xy$ , a student differentiates and obtains $\frac{\partial(u,v)}{\partial(x,y)}=x+y$ . Which critique is correct?

A.The result is correct because both variables contain

x+y

B.The determinant should be

x-y

, so the student's derivative setup is incomplete ✅

C.The determinant should be

y-x

, because the cross terms determine the sign

D.The Jacobian cannot be calculated because

v

contains a product

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: The partial derivatives are $u_x=1,\ u_y=1,\ v_x=y,\ v_y=x$ . Therefore the determinant is $1\cdot x-1\cdot y=x-y$ . The student's answer $x+y$ comes from confusing the derivatives of the product $xy$ with the original expression. This is a common differentiation error.

Q10. A graph shows a family of curves $u=\text{constant}$ forming parallel lines of slope $-1$ , while $v=\text{constant}$ forms parallel lines of slope $1$ . Which transformation is consistent with this geometry?

u=x+y,\ v=x-y

u=x-y,\ v=x+y

C.Both A and B ✅

u=x^2+y^2,\ v=xy

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: For $u=x+y$ , fixing $u$ gives $x+y=c$ , or $y=-x+c$ , a family of slope $-1$ lines. For $v=x-y$ , fixing $v$ gives $y=x-c$ , a family of slope $1$ lines. Reversing the names of the variables preserves the same two families, so both A and B fit the graph.

Q11. A region is described by $x^2+y^2\le9$ and lies entirely in the first quadrant. A student chooses $u=x^2+y^2$ and $v=\tan^{-1}(y/x)$ . Which advantage does this choice provide?

A.Both boundaries automatically become straight lines in the

xy

-plane

B.The circular and angular constraints become simple bounds in

u

and

v

, making the region easier to describe ✅

C.The Jacobian becomes zero everywhere

D.The transformation eliminates the need for an area-scaling factor

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The chosen variables represent radial distance squared and angular position. The circular boundary becomes $u\le9$ , while the first-quadrant restriction gives a simple angular interval for $v$ . This transforms a curved region into a rectangular-type parameter region, although the Jacobian is still essential for correctly transforming area elements.

Q12. Two students use different valid changes of variables to evaluate the same double integral. Student A obtains a simple rectangular region but a complicated Jacobian. Student B obtains a less simple region but a constant Jacobian. Which principle should determine the better method?

A.Always choose the transformation with the simplest Jacobian

B.Always choose the transformation with the simplest original integrand

C.Choose the transformation that makes the overall integral easiest, considering the integrand, region, and Jacobian together ✅

D.The two methods must produce different answers

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: A change of variables should be judged by the complete transformed integral, not by one feature alone. A simple region may compensate for a complicated Jacobian, while a constant Jacobian may compensate for more complicated bounds. The best transformation is the one that minimizes the combined difficulty of the integrand, bounds, and scaling factor.

Q13. Let $u=x^2-y^2$ and $v=2xy$ . At $(x,y)=(1,2)$ , what does the nonzero Jacobian tell you about the transformation near that point?

A.It guarantees the transformation is globally one-to-one

B.It indicates that the transformation preserves area exactly

C.It indicates that the transformation is locally invertible and does not collapse nearby two-dimensional area to first order ✅

D.It means the transformation maps every nearby point to the same location

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: The Jacobian at $(1,2)$ is nonzero because $4x^2+4y^2=20$ . A nonzero determinant means the transformation has locally independent directions and therefore is locally invertible near that point. It does not establish global one-to-one behavior, nor does it imply that area is preserved without scaling.

Q14. A transformation maps a small square of area $0.01$ near a point in the $uv$ -plane to a region whose area is approximately $0.04$ in the $xy$ -plane. Which conclusion is most reasonable about the local Jacobian magnitude?

A.Approximately

0.25

B.Approximately

0.5

C.Approximately

2

D.Approximately

4

✅

💡 Difficulty: medium | ✅ Correct: D

📖 Explanation: For sufficiently small regions, the absolute value of the Jacobian approximates the local area-scaling factor. The observed area changes from $0.01$ to approximately $0.04$ , so the scaling factor is $0.04/0.01=4$ . Thus the magnitude of $\frac{\partial(x,y)}{\partial(u,v)}$ is approximately $4$ near that point.

Q15. Suppose $u=x+y$ and $v=x-y$ . An integral is transformed using $dx\,dy=\frac12\,du\,dv$ . A student argues that the factor should be $2$ because $\frac{\partial(u,v)}{\partial(x,y)}=2$ . Which response best resolves the disagreement?

A.The student is correct because the Jacobian must always be multiplied

B.The student is incorrect because the area element requires the reciprocal Jacobian when expressed in terms of

du\,dv

✅

C.Both factors are correct simultaneously for the same direction

D.The Jacobian is irrelevant because the transformation is linear

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The determinant $\frac{\partial(u,v)}{\partial(x,y)}$ equals $-2$ , whose magnitude is $2$ . However, when rewriting $dx\,dy$ in terms of $du\,dv$ , the required factor is the absolute value of the inverse Jacobian, namely $\frac{1}{2}$ . The student's mistake is using the Jacobian in the wrong direction.

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