Jacobian Determinants - Multivariable Calculus

Let $T_1$ be the transformation from the $uv$ -plane to the $xy$ -plane defined by $x=u^2$ and $y=v$ . Let $T_2$ be a transformation from the $xy$ -plane to the $st$ -plane with Jacobian determinant $\frac{\partial(s,t)}{\partial(x,y)} = 3x$ . What is the Jacobian determinant of the composite transformation $T = T_2 \circ T_1$ , which maps from the $uv$ -plane to the $st$ -plane?
Answer: $6u^3$ The Jacobian of a composite transformation is the product of the Jacobians of the individual transformations, evaluated at the appropriate points. The chain rule for Jacobians states: $\frac{\partial(s,t)}{\partial(u,v)} = \frac{\partial(s,t)}{\partial(x,y)} \cdot \frac{\partial(x,y)}{\partial(u,v)}.$ First, we compute the Jacobian of $T_1$ : $J_1 = \frac{\partial(x,y)}{\partial(u,v)} = \det \begin{pmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{pmatrix} = \det \begin{pmatrix} 2u & 0 \\ 0 & 1 \end{pmatrix} = 2u.$ We are given the Jacobian of $T_2$ as $J_2 = \frac{\partial(s,t)}{\partial(x,y)} = 3x$ . To find the composite Jacobian in terms of $u$ and $v$ , we must substitute for $x$ : $J_2 = 3(u^2) = 3u^2$ . The Jacobian of the composite transformation is $J_T = J_2(x(u,v), y(u,v)) \cdot J_1(u,v) = (3x) \cdot (2u)$ . Substituting $x=u^2$ , we get $J_T = (3u^2)(2u) = 6u^3$ .
Consider the transformation from the $uv$ -plane to the $xy$ -plane given by $x = u^2 - v^2$ and $y = 2uv$ . Let $S$ be the square region in the $uv$ -plane defined by $1 \le u \le 2$ and $0 \le v \le 1$ . What is the area of the image of $S$ in the $xy$ -plane?
Answer: $\frac{32}{3}$ When you encounter a transformation between coordinate systems, the local area scaling factor is found using the Jacobian determinant. This measures how much the transformation stretches or compresses area elements at each point. For the transformation $x = e^u \cos v$ and $y = e^u \sin v$ , you need to compute the Jacobian matrix. Taking partial derivatives: $\frac{\partial x}{\partial u} = e^u \cos v$ , $\frac{\partial x}{\partial v} = -e^u \sin v$ $\frac{\partial y}{\partial u} = e^u \sin v$ , $\frac{\partial y}{\partial v} = e^u \cos v$ The Jacobian determinant is: $J = \frac{\partial x}{\partial u} \cdot \frac{\partial y}{\partial v} - \frac{\partial x}{\partial v} \cdot \frac{\partial y}{\partial u}$ $J = (e^u \cos v)(e^u \cos v) - (-e^u \sin v)(e^u \sin v) = e^{2u}(\cos^2 v + \sin^2 v) = e^{2u}$ At the point $(u,v) = (\ln 2, \pi/4)$ , we have $J = e^{2\ln 2} = e^{\ln 4} = 4$ , confirming answer D. Answer A (0) would mean the transformation is degenerate, which isn't true here. Answer B (2) might come from incorrectly calculating $e^{\ln 2} = 2$ instead of $e^{2\ln 2}$ . Answer C ( $2\sqrt{2}$ ) could result from mistakenly involving the specific $v$ -coordinate $\pi/4$ in the final calculation, when it actually cancels out in the Jacobian. Remember: for area scaling factors, always compute the absolute value of the Jacobian determinant. The specific point matters only for evaluating the final expression, not for the general form of the Jacobian.
Consider the affine transformation $T(u,v) = (3u - v + 1, u + 2v - 2)$ . Let $S$ be a circular disk in the $uv$ -plane with an area of $4\pi$ . What is the area of the image of $S$ under the transformation $T$ ?
Answer: $28\pi$ When you encounter area transformations in multivariable calculus, the key tool is the Jacobian determinant, which tells you how areas scale under the transformation. For the affine transformation $T(u,v) = (3u - v + 1, u + 2v - 2)$ , you need to find the Jacobian matrix by taking partial derivatives. If $x = 3u - v + 1$ and $y = u + 2v - 2$ , then: $J = \begin{pmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{pmatrix} = \begin{pmatrix} 3 & -1 \\ 1 & 2 \end{pmatrix}$ The determinant is $\det(J) = (3)(2) - (-1)(1) = 6 + 1 = 7$ . Under any linear transformation, areas are multiplied by $|\det(J)|$ . Since the original disk has area $4\pi$ , the transformed region has area $4\pi \cdot |7| = 28\pi$ . Answer A ( $4\pi$ ) assumes the transformation preserves area, ignoring the scaling effect. Answer B ( $20\pi$ ) likely comes from miscalculating the determinant as 5 instead of 7. Answer C ( $4\pi/7$ ) incorrectly divides by the determinant rather than multiplying—a common error when students confuse forward and inverse transformations. The correct answer is D: $28\pi$ . Study tip: For area transformation problems, always compute the Jacobian determinant and multiply the original area by its absolute value. The constant terms in affine transformations don't affect the determinant—only the linear coefficients matter for scaling.
The integral $\iint_R (x^2 + y^2) \, dx \, dy$ is transformed using $x = u+v$ and $y = u-v$ into the form $\iint_S g(u,v) \, du \, dv$ . What is the function $g(u,v)$ ?
Answer: $4(u^2+v^2)$ The function $g(u,v)$ is the product of the transformed integrand and the absolute value of the Jacobian determinant. First, transform the integrand: $x^2+y^2 = (u+v)^2 + (u-v)^2 = (u^2+2uv+v^2) + (u^2-2uv+v^2) = 2u^2+2v^2.$ Next, compute the Jacobian determinant $\frac{\partial(x,y)}{\partial(u,v)}$ : $J(u,v) = \det \begin{pmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{pmatrix} = \det \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix} = (1)(-1) - (1)(1) = -2.$ The absolute value of the Jacobian is $|-2|=2$ . Finally, multiply the transformed integrand by the absolute value of the Jacobian: $g(u,v) = (2u^2+2v^2) \cdot |J| = (2u^2+2v^2) \cdot 2 = 4(u^2+v^2).$
To evaluate an integral over the region $R$ in the $xy$ -plane bounded by the lines $y=x$ , $y=x+2$ , $y=-x+1$ , and $y=-x+3$ , a student proposes a change of variables $u=y-x$ and $v=y+x$ . What is the absolute value of the Jacobian determinant $|\frac{\partial(x,y)}{\partial(u,v)}|$ associated with this transformation?
Answer: $\frac{1}{2}$ When you encounter a change of variables problem in multivariable calculus, you need to find the Jacobian determinant to properly transform the integral. The Jacobian accounts for how area elements change under the transformation. Given the transformation $u = y - x$ and $v = y + x$ , you first need to solve for $x$ and $y$ in terms of $u$ and $v$ . Adding the equations: $u + v = 2y$ , so $y = \frac{u + v}{2}$ . Subtracting: $v - u = 2x$ , so $x = \frac{v - u}{2}$ . Now compute the Jacobian matrix $\frac{\partial(x,y)}{\partial(u,v)}$ : $\frac{\partial x}{\partial u} = -\frac{1}{2}, \quad \frac{\partial x}{\partial v} = \frac{1}{2}$ $\frac{\partial y}{\partial u} = \frac{1}{2}, \quad \frac{\partial y}{\partial v} = \frac{1}{2}$ The Jacobian determinant is: $\begin{vmatrix} -\frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & \frac{1}{2} \end{vmatrix} = (-\frac{1}{2})(\frac{1}{2}) - (\frac{1}{2})(\frac{1}{2}) = -\frac{1}{4} - \frac{1}{4} = -\frac{1}{2}$ Taking the absolute value gives $\frac{1}{2}$ , which is answer D. Answer A (2) likely comes from confusing this with the inverse Jacobian $\left|\frac{\partial(u,v)}{\partial(x,y)}\right|$ . Answer B (-2) makes the same error while forgetting the absolute value. Answer C (- $\frac{1}{2}$ ) is the correct determinant but without taking the absolute value. Remember: always take the absolute value of the Jacobian determinant when changing variables in integration, since area elements are always positive.
Consider the transformation $T$ from the $st$ -plane to the $xy$ -plane given by $x = s^2 - t^2$ and $y = 2st$ . This transformation fails to be one-to-one at certain points where the Jacobian determinant vanishes. At which of the following points does the transformation have a zero Jacobian?
Answer: $(s,t) = (0,0)$ The Jacobian matrix is $J = \begin{pmatrix} \frac{\partial x}{\partial s} & \frac{\partial x}{\partial t} \\ \frac{\partial y}{\partial s} & \frac{\partial y}{\partial t} \end{pmatrix} = \begin{pmatrix} 2s & -2t \\ 2t & 2s \end{pmatrix}$ . The determinant is $\det(J) = (2s)(2s) - (-2t)(2t) = 4s^2 + 4t^2 = 4(s^2 + t^2)$ . For the Jacobian to be zero, we need $s^2 + t^2 = 0$ , which occurs only when $s = 0$ and $t = 0$ . Let's check the given options: A) At $(1,1)$ : $\det(J) = 4(1 + 1) = 8 \neq 0$ . B) At $(0,1)$ : $\det(J) = 4(0 + 1) = 4 \neq 0$ . C) At $(1,0)$ : $\det(J) = 4(1 + 0) = 4 \neq 0$ . D) At $(0,0)$ : $\det(J) = 4(0 + 0) = 0$ . Therefore, the transformation has a zero Jacobian only at the origin. Choices A, B, and C represent common points students might check, but none of them yield a zero determinant.
For the transformation $u = x + y$ , $v = xy$ , the region $R$ in the $xy$ -plane bounded by $y = x$ , $y = 2x$ , $xy = 1$ , and $xy = 4$ is mapped to a rectangular region in the $uv$ -plane. To convert the double integral $\iint_R f(x,y) \, dx \, dy$ to $uv$ -coordinates, we need the Jacobian $\frac{\partial(x,y)}{\partial(u,v)}$ . If this Jacobian equals $\frac{1}{|u + 2\sqrt{v}|}$ , what is the correct form of the transformed integral?
Answer: $\iint_{R'} f\left(\frac{u + \sqrt{u^2 - 4v}}{2}, \frac{u - \sqrt{u^2 - 4v}}{2}\right) \frac{1}{|u + 2\sqrt{v}|} \, du \, dv$ From the system $u = x + y$ and $v = xy$ , we solve for $x$ and $y$ . Substituting $y = u - x$ into the second equation: $v = x(u - x) = ux - x^2$ . Rearranging: $x^2 - ux + v = 0$ . By the quadratic formula: $x = \frac{u \pm \sqrt{u^2 - 4v}}{2}$ . Since $y = u - x$ , we get $y = \frac{u \mp \sqrt{u^2 - 4v}}{2}$ . The question states that the Jacobian is $\frac{\partial(x,y)}{\partial(u,v)} = \frac{1}{|u + 2\sqrt{v}|}$ . When transforming integrals, we use: $\iint_R f(x,y) \, dx \, dy = \iint_{R'} f(x(u,v), y(u,v)) \left|\frac{\partial(x,y)}{\partial(u,v)}\right| \, du \, dv$ . The correct form uses the given Jacobian magnitude $\frac{1}{|u + 2\sqrt{v}|}$ with the appropriate expressions for $x$ and $y$ .
Consider the coordinate transformation $u = xy$ , $v = \frac{y}{x}$ from the $xy$ -plane to the $uv$ -plane. What is the absolute value of the Jacobian determinant $\left|\frac{\partial(u,v)}{\partial(x,y)}\right|$ in terms of $x$ and $y$ ?
Answer: $\frac{2|y|}{|x|}$ We compute the Jacobian matrix: $J = \begin{pmatrix} \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} \\ \frac{\partial v}{\partial x} & \frac{\partial v}{\partial y} \end{pmatrix} = \begin{pmatrix} y & x \\ -\frac{y}{x^2} & \frac{1}{x} \end{pmatrix}$ . The determinant is $\det(J) = y \cdot \frac{1}{x} - x \cdot \left(-\frac{y}{x^2}\right) = \frac{y}{x} + \frac{xy}{x^2} = \frac{y}{x} + \frac{y}{x} = \frac{2y}{x}$ . Therefore, $\left|\frac{\partial(u,v)}{\partial(x,y)}\right| = \left|\frac{2y}{x}\right| = \frac{2|y|}{|x|}$ . Choice A gives $\frac{2y}{x}$ without absolute value bars, which could be negative. Choice C is missing the factor of 2. Choice D gives $\frac{y^2}{x^2}$ , which would result from incorrectly squaring terms in the determinant calculation.
Consider the transformation from the $uv$ -plane to the $xy$ -plane given by $x = e^u \cos v$ and $y = e^u \sin v$ . What is the local area scaling factor of this transformation at the point $(u,v) = (\ln 2, \pi/4)$ ?
Answer: $4$ When you encounter a transformation between coordinate systems, the local area scaling factor is found using the Jacobian determinant. This measures how much the transformation stretches or compresses area elements at each point. For the transformation $x = e^u \cos v$ and $y = e^u \sin v$ , you need to compute the Jacobian matrix. Taking partial derivatives: $\frac{\partial x}{\partial u} = e^u \cos v$ , $\frac{\partial x}{\partial v} = -e^u \sin v$ $\frac{\partial y}{\partial u} = e^u \sin v$ , $\frac{\partial y}{\partial v} = e^u \cos v$ The Jacobian determinant is: $J = \frac{\partial x}{\partial u} \cdot \frac{\partial y}{\partial v} - \frac{\partial x}{\partial v} \cdot \frac{\partial y}{\partial u}$ $J = (e^u \cos v)(e^u \cos v) - (-e^u \sin v)(e^u \sin v) = e^{2u}(\cos^2 v + \sin^2 v) = e^{2u}$ At the point $(u,v) = (\ln 2, \pi/4)$ , we have $J = e^{2\ln 2} = e^{\ln 4} = 4$ , confirming answer D. Answer A (0) would mean the transformation is degenerate, which isn't true here. Answer B (2) might come from incorrectly calculating $e^{\ln 2} = 2$ instead of $e^{2\ln 2}$ . Answer C ( $2\sqrt{2}$ ) could result from mistakenly involving the specific $v$ -coordinate $\pi/4$ in the final calculation, when it actually cancels out in the Jacobian. Remember: for area scaling factors, always compute the absolute value of the Jacobian determinant. The specific point matters only for evaluating the final expression, not for the general form of the Jacobian.
Consider the transformation $T: (s,t) \mapsto (x,y)$ defined by $x = s + t$ , $y = s - t$ . If $\iint_S f(x,y) \, dx \, dy = \iint_R f(s+t, s-t) \cdot |J| \, ds \, dt$ , where $J$ is the Jacobian determinant $\frac{\partial(x,y)}{\partial(s,t)}$ , what is the value of $|J|$ ?
Answer: $2$ The Jacobian matrix is $J = \begin{pmatrix} \frac{\partial x}{\partial s} & \frac{\partial x}{\partial t} \\ \frac{\partial y}{\partial s} & \frac{\partial y}{\partial t} \end{pmatrix} = \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}$ . The determinant is $\det(J) = 1 \cdot (-1) - 1 \cdot 1 = -1 - 1 = -2$ . Therefore, $|J| = |-2| = 2$ . Choice A would result from incorrectly computing the determinant as $1 \cdot (-1) + 1 \cdot 1 = 0$ , then taking absolute value of something else. Choice C would result from confusing this with the inverse transformation or incorrectly computing $1/|J|$ . Choice D would result from incorrectly squaring the determinant.
The elliptical coordinates $(u,v)$ are related to Cartesian coordinates by $x = a u \cos v$ and $y = b u \sin v$ , where $a$ and $b$ are positive constants. At the point where $u = 2$ and $v = \frac{\pi}{3}$ , what is the Jacobian determinant $\frac{\partial(x,y)}{\partial(u,v)}$ ?
Answer: $2ab$ When you encounter coordinate transformations in multivariable calculus, the Jacobian determinant measures how area elements scale between coordinate systems. For the transformation $x = au\cos v$ and $y = bu\sin v$ , you need to find $\frac{\partial(x,y)}{\partial(u,v)}$ . Start by computing all four partial derivatives. For the first row: $\frac{\partial x}{\partial u} = a\cos v$ and $\frac{\partial x}{\partial v} = -au\sin v$ . For the second row: $\frac{\partial y}{\partial u} = b\sin v$ and $\frac{\partial y}{\partial v} = bu\cos v$ . The Jacobian determinant is: $\frac{\partial(x,y)}{\partial(u,v)} = \begin{vmatrix} a\cos v & -au\sin v \\ b\sin v & bu\cos v \end{vmatrix}$ $= (a\cos v)(bu\cos v) - (-au\sin v)(b\sin v) = abu\cos^2 v + abu\sin^2 v = abu(\cos^2 v + \sin^2 v) = abu$ At the point where $u = 2$ and $v = \frac{\pi}{3}$ , the Jacobian determinant equals $ab \cdot 2 = 2ab$ . Choice A ( $ab$ ) represents the Jacobian when $u = 1$ , missing the factor of 2. Choice B ( $4ab$ ) likely comes from incorrectly squaring the $u$ value. Choice D ( $ab\sqrt{3}$ ) suggests confusion about the trigonometric identity—you might get $\sqrt{3}$ if you incorrectly evaluate the trig functions at $v = \frac{\pi}{3}$ , but the identity $\cos^2 v + \sin^2 v = 1$ eliminates all trigonometric dependence. The correct answer is C. Remember: Jacobian determinants for coordinate transformations often simplify due to trigonometric identities, so look for ways to use $\cos^2 v + \sin^2 v = 1$ .
A transformation $T(u,v) = (x(u,v), y(u,v))$ has a Jacobian determinant $J(u,v) = \frac{\partial(x,y)}{\partial(u,v)}$ that is equal to zero for all points on the curve $u=v^2$ . Which of the following is the most accurate conclusion that can be drawn?
Answer: The transformation is not locally one-to-one at any point on the curve $u=v^2$ . The Jacobian determinant being zero at a point indicates that the transformation is singular at that point. By the inverse function theorem, if the Jacobian determinant is zero at a point, the function is not locally invertible (i.e., not locally one-to-one) at that point. This means that small areas around points on the curve $u=v^2$ are collapsed into lower-dimensional objects (curves or points). Distractor A is incorrect because regions not on the curve can have a non-zero area image. Distractor B is too strong; only the image of the curve $u=v^2$ is guaranteed to be degenerate. Distractor D is incorrect because a zero Jacobian is the boundary between orientation-preserving ( $J>0$ ) and orientation-reversing ( $J<0$ ); it does not itself reverse orientation.
A transformation from the $uvw$ -space to the $xyz$ -space is given by $x = 2u$ , $y = u+v$ , and $z=v-w^2$ . What is the absolute value of the Jacobian determinant $\frac{\partial(x,y,z)}{\partial(u,v,w)}$ at the point $(u,v,w) = (1, 2, 3)$ ?
Answer: 12 First, we compute the Jacobian matrix of the transformation: $J = \begin{pmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} & \frac{\partial x}{\partial w} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} & \frac{\partial y}{\partial w} \\ \frac{\partial z}{\partial u} & \frac{\partial z}{\partial v} & \frac{\partial z}{\partial w} \end{pmatrix} = \begin{pmatrix} 2 & 0 & 0 \\ 1 & 1 & 0 \\ 0 & 1 & -2w \end{pmatrix}$ The determinant of this lower-triangular matrix is the product of its diagonal entries: $\det(J) = (2)(1)(-2w) = -4w.$ Now, we evaluate this determinant at the point $(u,v,w) = (1,2,3)$ : $\det(J)|_{(1,2,3)} = -4(3) = -12.$ The question asks for the absolute value of the Jacobian determinant, which is $|-12|=12$ .
The hyperbolic coordinate system is defined by the transformation $x = u \cosh v$ and $y = u \sinh v$ , for $u > 0$ . What is the Jacobian determinant $\frac{\partial(x,y)}{\partial(u,v)}$ for this transformation? (Recall that $\cosh^2 v - \sinh^2 v = 1$ ).
Answer: $u$ To find the Jacobian determinant, we first compute the partial derivatives of $x$ and $y$ with respect to $u$ and $v$ : $\frac{\partial x}{\partial u} = \cosh v, \quad \frac{\partial x}{\partial v} = u \sinh v$ $\frac{\partial y}{\partial u} = \sinh v, \quad \frac{\partial y}{\partial v} = u \cosh v$ Now, we form the Jacobian matrix and compute its determinant: $J(u,v) = \det \begin{pmatrix} \cosh v & u \sinh v \\ \sinh v & u \cosh v \end{pmatrix} = (\cosh v)(u \cosh v) - (u \sinh v)(\sinh v)$ $= u \cosh^2 v - u \sinh^2 v = u(\cosh^2 v - \sinh^2 v)$ Using the provided identity $\cosh^2 v - \sinh^2 v = 1$ , the determinant simplifies to: $J(u,v) = u(1) = u.$
Let a coordinate transformation be defined by $u = 2x - 3y$ and $v = x + 4y$ . To evaluate an integral $\iint_R f(x,y) \, dA$ over a region $R$ in the $xy$ -plane, one changes variables to $u$ and $v$ . What is the correct expression for the area element $dA$ in terms of $du$ and $dv$ ?
Answer: $dA = \frac{1}{11} \, du \, dv$ The change of variables formula for a double integral requires the Jacobian determinant of the transformation from $(u,v)$ to $(x,y)$ . We are given $u(x,y)$ and $v(x,y)$ . The Jacobian of the inverse transformation is easier to compute first: $\frac{\partial(u,v)}{\partial(x,y)} = \det \begin{pmatrix} \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} \\ \frac{\partial v}{\partial x} & \frac{\partial v}{\partial y} \end{pmatrix} = \det \begin{pmatrix} 2 & -3 \\ 1 & 4 \end{pmatrix} = (2)(4) - (-3)(1) = 8 + 3 = 11.$ The Jacobian we need is the inverse of this one: $\frac{\partial(x,y)}{\partial(u,v)} = \left(\frac{\partial(u,v)}{\partial(x,y)}\right)^{-1} = \frac{1}{11}.$ The area element is then $dA = \left| \frac{\partial(x,y)}{\partial(u,v)} \right| \, du \, dv = \left| \frac{1}{11} \right| \, du \, dv = \frac{1}{11} \, du \, dv$ .
An integral is given by $\iint_R (x+y)^2 e^{x-y} \, dA$ , where $R$ is the parallelogram in the $xy$ -plane with vertices at $(1,0), (2,1), (1,2),$ and $(0,1)$ . Using the transformation $u=x+y$ and $v=x-y$ , which of the following integrals is equivalent to the original?
Answer: $\int_{-1}^1 \int_1^3 \frac{1}{2} u^2 e^v \, du \, dv$ First, transform the region. The vertices of $R$ in the $uv$ -plane are: $(1,0) \to (u=1, v=1)$ ; $(2,1) \to (u=3, v=1)$ ; $(1,2) \to (u=3, v=-1)$ ; $(0,1) \to (u=1, v=-1)$ . This is a rectangle $S$ defined by $1 \le u \le 3$ and $-1 \le v \le 1$ . The integrand becomes $u^2 e^v$ . Next, find the Jacobian $\frac{\partial(x,y)}{\partial(u,v)}$ . It's easier to find $\frac{\partial(u,v)}{\partial(x,y)} = \det \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix} = -2$ . Then, $\frac{\partial(x,y)}{\partial(u,v)} = (-2)^{-1} = -1/2$ . The area element is $dA = |-\frac{1}{2}| \, du \, dv = \frac{1}{2} \, du \, dv$ . The transformed integral is $\iint_S (u^2 e^v) \frac{1}{2} \, du \, dv$ , which is $\int_{-1}^1 \int_1^3 \frac{1}{2} u^2 e^v \, du \, dv$ .

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