Surface Parameterization & Area - Multivariable Calculus

A surface is described by the vector parameterization $\vec{r}(u,v) = \langle u \cos v, u \sin v, v \rangle$ for $0 \le u \le 2$ and $0 \le v \le \pi$ . This surface is a portion of a helicoid. What is its surface area element, $dS$ ?
Answer: $dS = \sqrt{1+u^2} \,du \,dv$ To find the surface area element $dS = ||\vec{r}_u \times \vec{r}_v|| \,du \,dv$ , we first compute the partial derivatives of $\vec{r}(u,v)$ . We have $\vec{r}_u = \langle \cos v, \sin v, 0 \rangle$ and $\vec{r}_v = \langle -u \sin v, u \cos v, 1 \rangle$ . Next, we compute their cross product: $\vec{r}_u \times \vec{r}_v = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \cos v & \sin v & 0 \\ -u \sin v & u \cos v & 1 \end{vmatrix} = \hat{i}(\sin v - 0) - \hat{j}(\cos v - 0) + \hat{k}(u \cos^2 v + u \sin^2 v) = \langle \sin v, -\cos v, u \rangle$ . Finally, we find the magnitude of this vector: $||\vec{r}_u \times \vec{r}_v|| = \sqrt{(\sin v)^2 + (-\cos v)^2 + u^2} = \sqrt{\sin^2 v + \cos^2 v + u^2} = \sqrt{1+u^2}$ . Therefore, the surface area element is $dS = \sqrt{1+u^2} \,du \,dv$ .
Consider the surface defined by $z = xy$ over the region $D: x^2 + y^2 \leq 4$ . If this surface is parametrized using $\mathbf{r}(x,y) = (x, y, xy)$ , what is the surface area element $dS$ ?
Answer: $\sqrt{1 + x^2 + y^2} \, dx \, dy$ For a surface $z = f(x,y)$ , the surface area element is $dS = \sqrt{1 + f_x^2 + f_y^2} \, dx \, dy$ . Here $f(x,y) = xy$ , so $f_x = y$ and $f_y = x$ . Therefore $dS = \sqrt{1 + y^2 + x^2} \, dx \, dy = \sqrt{1 + x^2 + y^2} \, dx \, dy$ . Choice B incorrectly includes an extra $x^2y^2$ term. Choice C uses the product $(f_x)(f_y) = xy$ instead of $f_x^2 + f_y^2$ . Choice D incorrectly uses $f_x^4 + f_y^4$ instead of $f_x^2 + f_y^2$ .
A surface $S$ is parameterized by $\vec{r}(u,v) = \langle u, v, 2uv \rangle$ . A new parameterization of $S$ is introduced with $u = s+t$ and $v = s-t$ , giving $\vec{R}(s,t)$ . What is the magnitude of the normal vector for the new parameterization, $||\vec{R}_s \times \vec{R}_t||$ ?
Answer: $2\sqrt{1+8s^2+8t^2}$ First, substitute the new parameters into the original parameterization: $\vec{R}(s,t) = \langle s+t, s-t, 2(s+t)(s-t) \rangle = \langle s+t, s-t, 2(s^2-t^2) \rangle$ . Now, compute the partial derivatives with respect to $s$ and $t$ : $\vec{R}_s = \langle 1, 1, 4s \rangle$ and $\vec{R}_t = \langle 1, -1, -4t \rangle$ . Next, compute the cross product: $\vec{R}_s \times \vec{R}_t = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 4s \\ 1 & -1 & -4t \end{vmatrix} = \hat{i}(-4t - (-4s)) - \hat{j}(-4t - 4s) + \hat{k}(-1 - 1) = \langle 4s-4t, 4s+4t, -2 \rangle$ . Finally, find the magnitude of this vector: $||\vec{R}_s \times \vec{R}_t|| = \sqrt{(4s-4t)^2 + (4s+4t)^2 + (-2)^2} = \sqrt{16(s^2-2st+t^2) + 16(s^2+2st+t^2) + 4} = \sqrt{16s^2-32st+16t^2 + 16s^2+32st+16t^2 + 4} = \sqrt{32s^2+32t^2+4} = \sqrt{4(1+8s^2+8t^2)} = 2\sqrt{1+8s^2+8t^2}$ .
The surface area element for a surface parameterized by $\vec{r}(u,v)$ is found to be $dS = \sqrt{1+4u^2+4v^2} \,du \,dv$ . Which of the following is most likely the surface being described, using $u$ and $v$ as parameters for the $x$ and $y$ coordinates, respectively?
Answer: The paraboloid $z = x^2+y^2$ We can test each surface by parameterizing it as $\vec{r}(u,v) = \langle u, v, f(u,v) \rangle$ and calculating its surface area element factor $\sqrt{1 + (f_u)^2 + (f_v)^2}$ . For choice B, the surface is $z=f(x,y)=x^2+y^2$ . We use parameters $u=x, v=y$ , so $f(u,v)=u^2+v^2$ . The partial derivatives are $f_u = 2u$ and $f_v = 2v$ . The magnitude of the normal vector is $\sqrt{1 + (2u)^2 + (2v)^2} = \sqrt{1+4u^2+4v^2}$ . This matches the given expression. For the other choices: A) Cone $z=\sqrt{u^2+v^2}$ gives $\sqrt{1 + (\frac{u}{\sqrt{u^2+v^2}})^2 + (\frac{v}{\sqrt{u^2+v^2}})^2} = \sqrt{2}$ . C) Plane $z=2u+2v$ gives $\sqrt{1+2^2+2^2}=\sqrt{9}=3$ . D) Hemisphere $z=\sqrt{1-u^2-v^2}$ gives a more complex expression $\frac{1}{\sqrt{1-u^2-v^2}}$ .
Let $S$ be the portion of the cylinder $x^2+z^2=9$ bounded by the planes $y=0$ and $y=4-x$ . The area of $S$ can be found by evaluating $\iint_D g(\theta,y) \,dy \,d\theta$ for some function $g$ and domain $D$ . Using the parameterization $x=3\cos\theta, z=3\sin\theta$ , which integral represents the area of $S$ ?
Answer: $\int_0^{2\pi} \int_0^{4-3\cos\theta} 3 \,dy \,d\theta$ First, we find the surface area element $dS$ . The parameterization is $\vec{r}(\theta, y) = \langle 3\cos\theta, y, 3\sin\theta \rangle$ . The partial derivatives are $\vec{r}_\theta = \langle -3\sin\theta, 0, 3\cos\theta \rangle$ and $\vec{r}_y = \langle 0, 1, 0 \rangle$ . Their cross product is $\vec{r}_\theta \times \vec{r}_y = \langle -3\cos\theta, 0, -3\sin\theta \rangle$ . The magnitude is $||\vec{r}_\theta \times \vec{r}_y|| = \sqrt{9\cos^2\theta + 9\sin^2\theta} = \sqrt{9} = 3$ . So, $dS = 3 \,dy \,d\theta$ . The bounds on $y$ are $0 \le y \le 4-x$ . Substituting $x=3\cos\theta$ , we get $0 \le y \le 4-3\cos\theta$ . Since the surface is not restricted to positive or negative $x$ values, the cylinder makes a full revolution, so the bounds for $\theta$ are $0 \le \theta \le 2\pi$ . The condition $y \ge 0$ means $4-3\cos\theta \ge 0$ , which is always true since the maximum of $3\cos\theta$ is 3. Thus, the integral is $\int_0^{2\pi} \int_0^{4-3\cos\theta} 3 \,dy \,d\theta$ .
A surface $S$ is parametrized by $\mathbf{r}(u,v) = (u\cos v, u\sin v, u^2)$ where $0 \leq u \leq 2$ and $0 \leq v \leq \pi$ . If the surface area element is $dS = ||\mathbf{r}_u \times \mathbf{r}_v|| \, du \, dv$ , which expression correctly represents $||\mathbf{r}_u \times \mathbf{r}_v||$ ?
Answer: $u\sqrt{1 + 4u^2}$ First find the partial derivatives: $\mathbf{r}_u = (\cos v, \sin v, 2u)$ and $\mathbf{r}_v = (-u\sin v, u\cos v, 0)$ . Then compute the cross product: $\mathbf{r}_u \times \mathbf{r}_v = (-2u^2\cos v, -2u^2\sin v, u)$ . The magnitude is $||\mathbf{r}_u \times \mathbf{r}_v|| = \sqrt{4u^4\cos^2 v + 4u^4\sin^2 v + u^2} = \sqrt{4u^4 + u^2} = u\sqrt{4u^2 + 1} = u\sqrt{1 + 4u^2}$ . Choice B incorrectly uses $2u^2$ instead of $4u^2$ . Choice C factors out $u$ incorrectly. Choice D omits the factor of $u$ entirely.
Consider the portion of the cylinder $x^2 + y^2 = 4$ between the planes $z = 0$ and $z = 3$ . If parametrized as $\mathbf{r}(\theta, z) = (2\cos\theta, 2\sin\theta, z)$ where $0 \leq \theta \leq 2\pi$ and $0 \leq z \leq 3$ , what is the surface area of this cylindrical surface?
Answer: $12\pi$ When you encounter surface area problems involving parametric surfaces, you need to use the formula $\text{Surface Area} = \iint_D \|\mathbf{r}_u \times \mathbf{r}_v\| \, du \, dv$ , where $\mathbf{r}_u$ and $\mathbf{r}_v$ are partial derivatives with respect to the parameters. For the parametrization $\mathbf{r}(\theta, z) = (2\cos\theta, 2\sin\theta, z)$ , let's find the partial derivatives: - $\mathbf{r}_\theta = (-2\sin\theta, 2\cos\theta, 0)$ - $\mathbf{r}_z = (0, 0, 1)$ The cross product is: $\mathbf{r}_\theta \times \mathbf{r}_z = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -2\sin\theta & 2\cos\theta & 0 \\ 0 & 0 & 1 \end{vmatrix} = (2\cos\theta, 2\sin\theta, 0)$ The magnitude is $\|\mathbf{r}_\theta \times \mathbf{r}_z\| = \sqrt{4\cos^2\theta + 4\sin^2\theta} = \sqrt{4} = 2$ . Therefore: $\text{Surface Area} = \int_0^{2\pi} \int_0^3 2 \, dz \, d\theta = \int_0^{2\pi} 6 \, d\theta = 12\pi$ Answer choice A ( $6\pi$ ) results from forgetting the factor of 2 in the radius when computing the cross product magnitude. Answer choice B ( $24\pi$ ) comes from incorrectly squaring the magnitude or doubling the final result. Answer choice D ( $36\pi$ ) likely results from using the wrong bounds or miscalculating the cross product entirely. Remember: for cylindrical surfaces, the cross product magnitude equals the radius of the cylinder. Always double-check your partial derivatives and cross product calculations—they're the most common sources of error in parametric surface area problems.
A cone is parametrized by $\mathbf{r}(u,v) = (u\cos v, u\sin v, 2u)$ where $1 \leq u \leq 3$ and $0 \leq v \leq 2\pi$ . What is the total surface area of this portion of the cone?
Answer: $8\pi\sqrt{5}$ When you encounter a parametric surface and need to find its area, you're working with the fundamental formula: surface area equals the double integral of the magnitude of the cross product of the partial derivatives of the position vector. Start by finding the partial derivatives of $\mathbf{r}(u,v) = (u\cos v, u\sin v, 2u)$ . You get $\mathbf{r}_u = (\cos v, \sin v, 2)$ and $\mathbf{r}_v = (-u\sin v, u\cos v, 0)$ . Next, compute their cross product: $\mathbf{r}_u \times \mathbf{r}_v = (-2u\cos v, -2u\sin v, u)$ . The magnitude is $|\mathbf{r}_u \times \mathbf{r}_v| = \sqrt{4u^2\cos^2 v + 4u^2\sin^2 v + u^2} = \sqrt{4u^2 + u^2} = u\sqrt{5}$ . The surface area integral becomes $\int_0^{2\pi} \int_1^3 u\sqrt{5} \, du \, dv = \sqrt{5} \int_0^{2\pi} \int_1^3 u \, du \, dv = \sqrt{5} \cdot 2\pi \cdot \left[\frac{u^2}{2}\right]_1^3 = \sqrt{5} \cdot 2\pi \cdot \frac{8}{2} = 8\pi\sqrt{5}$ . Choice A ( $16\pi\sqrt{5}$ ) likely comes from incorrectly calculating $\int_1^3 u \, du$ as 8 instead of 4. Choice B ( $4\pi\sqrt{5}$ ) probably results from using $2u$ instead of $u\sqrt{5}$ as the integrand. Choice C ( $12\pi\sqrt{5}$ ) might stem from computational errors in the cross product or integral evaluation. Remember: for parametric surfaces, always compute $|\mathbf{r}_u \times \mathbf{r}_v|$ carefully—the cross product magnitude often involves recognizing perfect squares under the radical.
A torus (donut shape) is generated by rotating a circle of radius $r$ centered at $(R,0,0)$ in the $xz$ -plane about the $z$ -axis, where $0 < r < R$ . A standard parameterization is $\vec{r}(\theta, \phi) = \langle (R+r\cos\phi)\cos\theta, (R+r\cos\phi)\sin\theta, r\sin\phi \rangle$ . What is the surface area element $dS$ for this torus?
Answer: $dS = r(R+r\cos\phi) \,d\theta \,d\phi$ When you encounter a parameterized surface, finding the surface area element requires computing the magnitude of the cross product of the partial derivatives. This fundamental technique appears frequently in multivariable calculus. For the torus parameterization $\vec{r}(\theta, \phi) = \langle (R+r\cos\phi)\cos\theta, (R+r\cos\phi)\sin\theta, r\sin\phi \rangle$ , you need to find $\vec{r}_\theta$ and $\vec{r}_\phi$ : $\vec{r}_\theta = \langle -(R+r\cos\phi)\sin\theta, (R+r\cos\phi)\cos\theta, 0 \rangle$ $\vec{r}_\phi = \langle -r\sin\phi\cos\theta, -r\sin\phi\sin\theta, r\cos\phi \rangle$ Computing their cross product $\vec{r}_\theta \times \vec{r}_\phi$ and finding its magnitude gives $|\vec{r}_\theta \times \vec{r}_\phi| = r(R+r\cos\phi)$ . Therefore, $dS = r(R+r\cos\phi) \,d\theta \,d\phi$ . Answer A omits the factor of $r$ , which would arise if you mistakenly used just the distance from the z-axis without accounting for the circle's radius contribution. Answer B incorrectly includes an extra factor of $R$ , perhaps from confusing this with a different geometric setup. Answer C gives a constant $rR$ , which ignores the varying distance $(R+r\cos\phi)$ from the z-axis as the generating circle moves around. Remember: for any parameterized surface, the surface area element always equals the magnitude of the cross product of partial derivatives. Don't try to shortcut this calculation—work through the cross product systematically to avoid missing important geometric factors.
A surface $S$ is parameterized by $\vec{r}(u,v)$ over a domain $D$ . The surface area is given by $A(S) = \iint_D ||\vec{r}_u \times \vec{r}_v|| \,du \,dv$ . Consider a re-parameterization given by $\vec{R}(s,t) = \vec{r}(s^2, t^2)$ . Using the chain rule, how does the new surface area scaling factor, $||\vec{R}_s \times \vec{R}_t||$ , relate to the original scaling factor, $||\vec{r}_u \times \vec{r}_v||$ ?
Answer: $||\vec{R}_s \times \vec{R}_t|| = 4st \, ||\vec{r}_u \times \vec{r}_v||$ We apply the chain rule for vector functions. Let $u(s,t) = s^2$ and $v(s,t) = t^2$ . Then $\vec{R}_s = \frac{\partial \vec{r}}{\partial u}\frac{\partial u}{\partial s} + \frac{\partial \vec{r}}{\partial v}\frac{\partial v}{\partial s} = \vec{r}_u (2s) + \vec{r}_v (0) = 2s \vec{r}_u$ . Similarly, $\vec{R}_t = \frac{\partial \vec{r}}{\partial u}\frac{\partial u}{\partial t} + \frac{\partial \vec{r}}{\partial v}\frac{\partial v}{\partial t} = \vec{r}_u (0) + \vec{r}_v (2t) = 2t \vec{r}_v$ . In both cases, the partials $\vec{r}_u$ and $\vec{r}_v$ are evaluated at $(u,v)=(s^2, t^2)$ . Now we compute the cross product: $\vec{R}_s \times \vec{R}_t = (2s \vec{r}_u) \times (2t \vec{r}_v) = (2s)(2t) (\vec{r}_u \times \vec{r}_v) = 4st (\vec{r}_u \times \vec{r}_v)$ . Taking the magnitude of both sides gives $||\vec{R}_s \times \vec{R}_t|| = |4st| \, ||\vec{r}_u \times \vec{r}_v||$ . Assuming $s, t$ are positive, this simplifies to $4st \, ||\vec{r}_u \times \vec{r}_v||$ .
The surface of a cone with height $H$ and radius $R$ is given by $z = H(1 - \frac{\sqrt{x^2+y^2}}{R})$ . A student parameterizes the cone using $\vec{r}(r,\theta) = \langle r\cos\theta, r\sin\theta, H(1-r/R) \rangle$ for $0 \le r \le R$ and $0 \le \theta \le 2\pi$ . Which of the following is the correct surface area element $dS$ for this parameterization?
Answer: $dS = \frac{\sqrt{R^2+H^2}}{R} r \,dr \,d\theta$ When calculating surface area using parametric equations, you need to find the magnitude of the cross product of the partial derivative vectors. This measures how the surface "stretches" compared to the parameter space. Given $\vec{r}(r,\theta) = \langle r\cos\theta, r\sin\theta, H(1-r/R) \rangle$ , you first compute the partial derivatives: $\vec{r}_r = \langle \cos\theta, \sin\theta, -H/R \rangle$ $\vec{r}_\theta = \langle -r\sin\theta, r\cos\theta, 0 \rangle$ Next, find their cross product: $\vec{r}_r \times \vec{r}_\theta = \langle \frac{Hr\cos\theta}{R}, \frac{Hr\sin\theta}{R}, r \rangle$ The magnitude is: $|\vec{r}_r \times \vec{r}_\theta| = \sqrt{\frac{H^2r^2\cos^2\theta}{R^2} + \frac{H^2r^2\sin^2\theta}{R^2} + r^2} = \sqrt{\frac{H^2r^2}{R^2} + r^2} = r\sqrt{\frac{H^2 + R^2}{R^2}} = \frac{r\sqrt{R^2+H^2}}{R}$ Therefore, $dS = \frac{\sqrt{R^2+H^2}}{R} r \,dr \,d\theta$ , which is answer D. Answer A incorrectly has $r^2$ under the square root instead of factoring out $r$ . Answer B uses $(1 + H/R)$ instead of the proper Pythagorean relationship and includes an extra $r$ factor. Answer C completely omits the $r$ term that comes from the Jacobian of the cylindrical coordinate transformation. Study tip: Remember that surface area elements always involve the magnitude of $\vec{r}_r \times \vec{r}_\theta$ . For surfaces of revolution or cones, expect to see $\sqrt{1 + (\text{slope})^2}$ patterns, and don't forget the $r$ factor when using cylindrical-like coordinates.
A surface of revolution is formed by rotating the curve $y=e^{x/2}$ for $0 \le x \le 2$ about the x-axis. Which of the following expressions represents the surface area element $dS$ for this surface, using parameters $x$ and $\theta$ (the angle of revolution)?
Answer: $dS = e^{x/2}\sqrt{1+\frac{1}{4}e^x} \,dx \,d\theta$ A parameterization for a surface of revolution of $y=f(x)$ about the x-axis is $\vec{r}(x,\theta) = \langle x, f(x)\cos\theta, f(x)\sin\theta \rangle$ . Here, $f(x)=e^{x/2}$ . So, $\vec{r}(x,\theta) = \langle x, e^{x/2}\cos\theta, e^{x/2}\sin\theta \rangle$ . The partial derivatives are $\vec{r}_x = \langle 1, \frac{1}{2}e^{x/2}\cos\theta, \frac{1}{2}e^{x/2}\sin\theta \rangle$ and $\vec{r}_\theta = \langle 0, -e^{x/2}\sin\theta, e^{x/2}\cos\theta \rangle$ . The cross product $\vec{r}_x \times \vec{r}_\theta$ is $\langle \frac{1}{2}e^x, -e^{x/2}\cos\theta, -e^{x/2}\sin\theta \rangle$ . The magnitude is $||\vec{r}_x \times \vec{r}_\theta|| = \sqrt{(\frac{1}{2}e^x)^2 + (-e^{x/2}\cos\theta)^2 + (-e^{x/2}\sin\theta)^2} = \sqrt{\frac{1}{4}e^{2x} + e^x(\cos^2\theta+\sin^2\theta)} = \sqrt{\frac{1}{4}e^{2x} + e^x} = \sqrt{e^x(\frac{1}{4}e^x+1)} = e^{x/2}\sqrt{1+\frac{1}{4}e^x}$ . Thus, $dS = e^{x/2}\sqrt{1+\frac{1}{4}e^x} \,dx \,d\theta$ .
The graph of $z = x^2 - y^2$ over the square region $[-1,1] \times [-1,1]$ forms a surface $S$ . If we use the standard parametrization $\mathbf{r}(x,y) = (x, y, x^2 - y^2)$ , what is the surface area element $dS$ in terms of $x$ and $y$ ?
Answer: $\sqrt{1 + 4x^2 + 4y^2} \, dx \, dy$ When you encounter a surface area problem in multivariable calculus, you need to find the surface area element $dS$ using the formula $dS = \|\mathbf{r}_x \times \mathbf{r}_y\| \, dx \, dy$ , where $\mathbf{r}_x$ and $\mathbf{r}_y$ are the partial derivatives of the parametrization vector. Given $\mathbf{r}(x,y) = (x, y, x^2 - y^2)$ , let's find the partial derivatives: - $\mathbf{r}_x = (1, 0, 2x)$ - $\mathbf{r}_y = (0, 1, -2y)$ Now compute the cross product $\mathbf{r}_x \times \mathbf{r}_y$ : $\mathbf{r}_x \times \mathbf{r}_y = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 0 & 2x \\ 0 & 1 & -2y \end{vmatrix} = (-2x, 2y, 1)$ The magnitude is: $\|\mathbf{r}_x \times \mathbf{r}_y\| = \sqrt{(-2x)^2 + (2y)^2 + 1^2} = \sqrt{4x^2 + 4y^2 + 1} = \sqrt{1 + 4x^2 + 4y^2}$ Therefore, $dS = \sqrt{1 + 4x^2 + 4y^2} \, dx \, dy$ , which is answer choice C. Answer A incorrectly subtracts the $y$ term, likely from confusing the original function $x^2 - y^2$ with the cross product calculation. Answer B uses coefficients of 2 instead of 4, suggesting an error in squaring the partial derivatives $2x$ and $-2y$ . Answer D omits the factor of 4 entirely, possibly from forgetting to square the partial derivatives. Remember: when computing surface area elements, always square each component of the cross product before adding them under the square root. The coefficients matter!
Let $S$ be the portion of the sphere $x^2+y^2+z^2=4$ that lies above the $xy$ -plane and within the cylinder $x^2+y^2=3$ . If $S$ is parameterized as a graph $z=f(x,y)$ , its surface area element is $dS = g(x,y) \,dx \,dy$ . What is $g(x,y)$ ?
Answer: $g(x,y) = \frac{2}{\sqrt{4-x^2-y^2}}$ The surface is the graph of $z = f(x,y) = \sqrt{4-x^2-y^2}$ . The surface area element is given by $dS = \sqrt{1 + (\frac{\partial z}{\partial x})^2 + (\frac{\partial z}{\partial y})^2} \,dx \,dy$ . We compute the partial derivatives: $\frac{\partial z}{\partial x} = \frac{-2x}{2\sqrt{4-x^2-y^2}} = \frac{-x}{\sqrt{4-x^2-y^2}}$ and $\frac{\partial z}{\partial y} = \frac{-y}{\sqrt{4-x^2-y^2}}$ . Squaring and adding them gives $(\frac{\partial z}{\partial x})^2 + (\frac{\partial z}{\partial y})^2 = \frac{x^2}{4-x^2-y^2} + \frac{y^2}{4-x^2-y^2} = \frac{x^2+y^2}{4-x^2-y^2}$ . Then, $1 + (\frac{\partial z}{\partial x})^2 + (\frac{\partial z}{\partial y})^2 = 1 + \frac{x^2+y^2}{4-x^2-y^2} = \frac{(4-x^2-y^2) + (x^2+y^2)}{4-x^2-y^2} = \frac{4}{4-x^2-y^2}$ . Taking the square root, we get $g(x,y) = \sqrt{\frac{4}{4-x^2-y^2}} = \frac{2}{\sqrt{4-x^2-y^2}}$ . While $g(x,y)=2/z$ is technically correct, standard practice requires expressing the function in terms of the parameters $x$ and $y$ .
The area of a small patch of a parameterized surface $\vec{r}(u,v)$ is approximated by the area of a parallelogram on its tangent plane, given by $dA_{surf} \approx ||\vec{r}_u \times \vec{r}_v|| \,du \,dv$ . What does the quantity $||\vec{r}_u \times \vec{r}_v||$ represent geometrically?
Answer: The local rate at which the parameterization stretches or shrinks area from the $uv$ -plane to the surface in $\mathbb{R}^3$ . The expression $||\vec{r}_u \times \vec{r}_v||$ is the surface area scaling factor, also known as the Jacobian for the surface parameterization. It describes how area is transformed from the parameter domain (the $uv$ -plane) to the surface in 3D space. A small rectangle in the $uv$ -plane with area $du \,dv$ is mapped to a small patch of the surface with area $dS = ||\vec{r}_u \times \vec{r}_v|| \,du \,dv$ . Therefore, $||\vec{r}_u \times \vec{r}_v||$ is the local ratio of the area on the surface to the corresponding area in the parameter domain, representing the local stretching or shrinking of area.
Consider two different parametrizations of the same surface: $\mathbf{r}_1(u,v) = (u, v, uv)$ and $\mathbf{r}_2(s,t) = (s^2, \frac{t}{2}, \frac{s^2 t}{2})$ where the domains are chosen so that both cover the same portion of the surface. If $||\mathbf{r}_{1u} \times \mathbf{r}_{1v}|| = \sqrt{1 + u^2 + v^2}$ at a point $(u_0, v_0)$ , and the corresponding point in the second parametrization is $(s_0, t_0)$ where $s_0^2 = u_0$ and $\frac{t_0}{2} = v_0$ , what is $||\mathbf{r}_{2s} \times \mathbf{r}_{2t}||$ at $(s_0, t_0)$ ?
Answer: $s_0\sqrt{1 + s_0^4 + \frac{t_0^2}{4}}$ When you encounter surface parametrizations, the key insight is understanding how different coordinate systems relate through the Jacobian of the transformation. The magnitude of the cross product of partial derivatives (which gives surface area elements) transforms according to this Jacobian determinant. First, let's establish the coordinate transformation. From the given relationships: $u = s^2$ , $v = \frac{t}{2}$ , and $w = uv = s^2 \cdot \frac{t}{2} = \frac{s^2t}{2}$ The Jacobian matrix of this transformation is: $J = \begin{pmatrix} \frac{\partial u}{\partial s} & \frac{\partial u}{\partial t} \\ \frac{\partial v}{\partial s} & \frac{\partial v}{\partial t} \end{pmatrix} = \begin{pmatrix} 2s & 0 \\ 0 & \frac{1}{2} \end{pmatrix}$ The determinant is $|J| = 2s \cdot \frac{1}{2} = s$ . For surface parametrizations, the cross product magnitudes are related by: $||\mathbf{r}_{2s} \times \mathbf{r}_{2t}|| = |J| \cdot ||\mathbf{r}_{1u} \times \mathbf{r}_{1v}||$ At the corresponding points, $u_0 = s_0^2$ and $v_0 = \frac{t_0}{2}$ , so: $||\mathbf{r}_{2s} \times \mathbf{r}_{2t}|| = s_0 \cdot \sqrt{1 + u_0^2 + v_0^2} = s_0\sqrt{1 + s_0^4 + \frac{t_0^2}{4}}$ This matches answer B. A incorrectly uses $\frac{s_0}{2}$ instead of $s_0$ as the Jacobian factor. C uses $2s_0$ , likely from confusing which transformation direction to use. D incorrectly manipulates the expression under the square root, possibly from algebraic errors in the substitution. Study tip: Always compute the Jacobian determinant carefully when transforming between parametrizations—the cross product magnitude scales by this factor.
A hemisphere of radius 3 is parametrized by $\mathbf{r}(\phi, \theta) = (3\sin\phi\cos\theta, 3\sin\phi\sin\theta, 3\cos\phi)$ where $0 \leq \phi \leq \frac{\pi}{2}$ and $0 \leq \theta \leq 2\pi$ . Which expression gives the correct magnitude $||\mathbf{r}_{\phi} \times \mathbf{r}_{\theta}||$ for the surface area element?
Answer: $9\sin\phi$ When you encounter a parametric surface and need to find the surface area element, you're looking for the magnitude of the cross product of the partial derivative vectors. This cross product gives you a vector normal to the surface, and its magnitude tells you how much the surface area is "stretched" at each point. To find $||\mathbf{r}_{\phi} \times \mathbf{r}_{\theta}||$ , you need to compute the partial derivatives first: $\mathbf{r}_{\phi} = (3\cos\phi\cos\theta, 3\cos\phi\sin\theta, -3\sin\phi)$ $\mathbf{r}_{\theta} = (-3\sin\phi\sin\theta, 3\sin\phi\cos\theta, 0)$ Computing their cross product: $\mathbf{r}_{\phi} \times \mathbf{r}_{\theta} = (9\sin^2\phi\cos\theta, 9\sin^2\phi\sin\theta, 9\sin\phi\cos\phi)$ The magnitude is: $||\mathbf{r}_{\phi} \times \mathbf{r}_{\theta}|| = \sqrt{81\sin^4\phi\cos^2\theta + 81\sin^4\phi\sin^2\theta + 81\sin^2\phi\cos^2\phi}$ Factoring out $81\sin^2\phi$ : $= 9\sin\phi\sqrt{\sin^2\phi(\cos^2\theta + \sin^2\theta) + \cos^2\phi} = 9\sin\phi\sqrt{\sin^2\phi + \cos^2\phi} = 9\sin\phi$ This confirms answer (B) is correct. (A) $3\sin\phi$ misses the factor of 3 that comes from squaring the radius. (C) $9\sin^2\phi$ appears if you forget to take the square root when computing the magnitude. (D) $9\cos\phi$ would result from confusing $\phi$ and $\theta$ or misremembering spherical coordinate identities. Study tip: For spherical coordinates, the surface area element always involves $\sin\phi$ (never $\cos\phi$ ), and remember that the radius gets squared in the cross product magnitude.

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