📝 Change of Variables Formula for Double Integrals (15 MCQs)

📖 Explanation: The Jacobian matrix is $\begin{pmatrix}1&1\\1&-1\end{pmatrix}$. Its determinant is $1(-1)-1(1)=-2$, so the absolute value is $2$. The absolute value is required because area scaling must be nonnegative. A common mistake is to use the signed determinant directly or confuse the determinant with its reciprocal.

📖 Explanation: The Jacobian determinant measures signed local area scaling, while area itself cannot be negative. Therefore, the absolute value gives the actual factor by which a small region in the $uv$-plane changes area in the $xy$-plane. It does not guarantee one-to-one behavior or eliminate the need to transform boundaries.

📖 Explanation: The Jacobian matrix is $\begin{pmatrix}2&0\\0&3\end{pmatrix}$, whose determinant is $6$. Thus every small area is multiplied by $6$. Since the original region has area $5$, the transformed area is $6(5)=30$. The result follows from local area scaling and does not require evaluating a double integral.

📖 Explanation: The determinant is $-2$, but the transformed area element is $dx\,dy=|-2|\,du\,dv=2\,du\,dv$. The negative sign indicates orientation reversal rather than negative physical area. Failing to take the absolute value can produce an incorrect sign for an otherwise positive area or integral.

📖 Explanation: Each boundary is already expressed using either $x+y$ or $x-y$. Setting $u=x+y$ and $v=x-y$ converts the four boundaries into $u=1$, $u=4$, $v=2$, and $v=5$, producing a rectangle in the $uv$-plane. This simplifies both the region and the integration limits.

📖 Explanation: The magnitude of the determinant gives the area scaling factor, so the local area is multiplied by $3$. The negative sign indicates that orientation is reversed. Therefore, the correct interpretation combines both pieces of information: magnitude $3$ for area scaling and a sign change for orientation.

📖 Explanation: The change of variables requires the area element to be multiplied by the absolute Jacobian. Since the Jacobian magnitude is $4$, the mass becomes $4\int_1^3\int_0^2(u+v)\,dv\,du$. The bounds describe the transformed rectangular region, so changing them or using the reciprocal Jacobian would produce an incorrect mass.

📖 Explanation: Solving for the inverse gives $x=(u+v)/2$ and $y=(u-v)/2$. Therefore, the inverse Jacobian has determinant $-1/2$, whose magnitude is $1/2$. Hence $dx\,dy=\frac12\,du\,dv$. The student likely computed the forward Jacobian, $2$, but used it in the wrong direction.

📖 Explanation: For a sufficiently small region, a differentiable transformation behaves approximately like its linearization. The determinant of its derivative therefore controls local area scaling. Here the relevant quantity is the absolute Jacobian determinant. The curved appearance of the image does not change this local principle; higher-order curvature effects become negligible for sufficiently small regions.

📖 Explanation: Each family of parallel boundary lines becomes a constant-coordinate boundary. The equations involving $x+y$ become $u=1$ and $u=3$, while those involving $x-y$ become $v=0$ and $v=2$. Thus the parallelogram becomes the rectangle $1\le u\le3$, $0\le v\le2$.

📖 Explanation: A useful transformation simplifies both the geometry and the differential area element. If the boundaries become coordinate lines and the Jacobian is constant and nonzero, the transformed integral often has much simpler limits and scaling. However, the integrand still must be transformed, and the resulting problem remains a double integral.

📖 Explanation: Within the region, $u-v$ is positive in some parts and negative in others. A signed Jacobian would therefore introduce artificial cancellation between areas. The correct area factor is $|u-v|$. In addition, a zero Jacobian occurs along $u=v$, so the transformation's local invertibility must be considered carefully.

📖 Explanation: The change-of-variables formula replaces $dx\,dy$ by the absolute Jacobian multiplied by $du\,dv$. Since the Jacobian magnitude is $5$, the transformed integral is 5\iint_{R'}g(u,v)\,du\,dv. The integrand must already be expressed in terms of $u,v$, so retaining $f(x,y)$ without substitution is incomplete.

📖 Explanation: A zero Jacobian means that the derivative matrix loses rank at that point, so local two-dimensional area can collapse in one direction. This prevents the usual local inverse interpretation there. It does not by itself prove that the entire transformation is invalid or that the original integral equals zero; additional global information is required.

📖 Explanation: The Jacobian matrix is $\begin{pmatrix}1&1\\1&-1\end{pmatrix}$, with determinant $-2$, so the area-scaling magnitude is $2$. The $uv$-rectangle has area $1$, hence its image has area $2$. The negative sign indicates orientation reversal and does not make the physical area negative. Thus the second claim is correct.