📝 Change of Variables in Multiple Integrals: Jacobians (14 MCQs)

📖 Explanation: A change of variables scales area according to the absolute value of the Jacobian determinant. Here the derivative matrix is $\begin{pmatrix}2&-1\\1&3\end{pmatrix}$, whose determinant is $7$. Therefore a small $du,dv$ region becomes an area approximately $7\,du\,dv$. The other choices do not correctly describe local area scaling.

📖 Explanation: Symmetry in the formulas does not imply unit area scaling. The derivative matrix is $\begin{pmatrix}1&1\\1&-1\end{pmatrix}$, so its determinant is $-2$. The absolute value, $2$, gives the area scaling factor. The negative sign describes orientation, not a negative physical area.

📖 Explanation: The Jacobian determinant can be negative when a transformation reverses orientation. However, geometric area must remain nonnegative, so the conversion uses the absolute value of the determinant. This distinction is important because the determinant contains both scaling information and orientation information, while an area element represents magnitude.

📖 Explanation: A zero Jacobian indicates that local area scaling collapses at that point. This often signals that the transformation is not locally invertible there, although it does not mean that the entire transformed region has zero area. The integrand itself also need not vanish. Local invertibility must therefore be examined carefully.

📖 Explanation: The transformation stretches distances in the $u$-direction by $3$ and in the $v$-direction by $2$. Therefore the area scaling factor is $3\cdot2=6$, which is also the absolute Jacobian determinant. Multiplying the original area $5$ by $6$ gives an image area of $30$, so none of the listed values is correct.

📖 Explanation: The forward Jacobian is $\frac{\partial(x,y)}{\partial(u,v)}=-2$. Since area uses the absolute value, $|J|=2$, giving $dx\,dy=2\,du\,dv$. The reciprocal value would correspond to the inverse Jacobian when expressing $du\,dv$ in terms of $dx\,dy$, which is a common source of mistakes.

📖 Explanation: The boundary curves already have the forms $x+y=\text{constant}$ and $x-y=\text{constant}$. Choosing $u=x+y$ and $v=x-y$ converts all four boundaries into constant-coordinate lines, producing a rectangle in the $uv$-plane. This substantially simplifies both the region description and the limits.

📖 Explanation: The derivative matrix is diagonal with entries $2u$ and $2v$, giving determinant $4uv$. For an area element, the required factor is $|4uv|=4|uv|$. Writing $4uv$ without absolute values can produce negative area factors when $u$ and $v$ have opposite signs, so the domain and one-to-one behavior must be considered.

📖 Explanation: The boundary curves are already described by constant products and ratios. Setting $u=xy$ changes the product boundaries to $u=1$ and $u=4$, while $v=x/y$ changes the ratio boundaries to $v=1$ and $v=9$. Thus the curved region becomes a rectangle in the transformed coordinates, making the geometry much easier to handle.

📖 Explanation: The identity $\cos^2\theta+\sin^2\theta=1$ is relevant to simplifying the determinant, but it is not itself the Jacobian. Differentiating $x$ and $y$ with respect to $r,\theta$ gives a determinant of $r$, so $dx\,dy=r\,dr\,d\theta$. The radial factor is essential for correct area scaling.

📖 Explanation: The determinant magnitude gives the local area scaling, so a determinant of $-3$ means the area is multiplied by $3$. The negative sign indicates orientation reversal. A small square therefore becomes a parallelogram with approximately three times its original area, assuming the transformation is sufficiently smooth near the point.

📖 Explanation: The derivative matrix is $\begin{pmatrix}1&2\\3&6\end{pmatrix}$. Its determinant is $1(6)-2(3)=0$, so the transformation collapses two-dimensional regions into lower-dimensional sets locally. A nonzero Jacobian calculation would indicate an algebraic error. Before changing variables, local invertibility and nonzero area scaling must be verified.

📖 Explanation: Differentiating gives $x_u=2u,\ x_v=-2v,\ y_u=2v,\ y_v=2u$. The determinant is $4u^2+4v^2=4(u^2+v^2)$. This is positive away from the origin, indicating nonzero local area scaling there. The result also shows that the scaling grows with the squared distance from the $uv$-origin.

📖 Explanation: The circular boundaries become constant values of $r$, while the rays become constant values of $\theta$. In addition, $x^2+y^2=r^2$, and the area element contributes another factor $r$. Thus the substitution simultaneously simplifies the region, the integrand, and the differential element, reducing a complicated two-dimensional problem to straightforward separated limits.