📝 Parametric representation of surfaces of revolution (13 MCQs)

📖 Explanation: The correct parametrization for revolving $z = y^2$ about the $z$-axis uses $x = v \cos u$, $y = v \sin u$, and $z = v^2$, where $u$ is the azimuthal angle and $v$ is the distance from the $z$-axis. Option A incorrectly assigns $z = v^2$ but swaps coordinates; option C gives a cone; option D distorts the radial component. Understanding parameter roles is key.

📖 Explanation: The student's parametrization is standard: $\mathbf{r}(u,v) = (f(u)\cos v, f(u)\sin v, u)$. The cross product magnitude yields |f(u)| \sqrt{1+[f'(u)]^2}, so the student's claim is actually correct. The error analysis must recognize that no mistake exists; other options misrepresent the Jacobian or the radius factor.

📖 Explanation: The torus parametrization uses $u$ for the circular cross-section and $v$ for the revolution angle. Option A is standard with $u,v \in [0,2\pi]$. Option B uses sine/cosine swapped incorrectly; option C is planar; option D has mismatched trigonometric arguments. The domain must cover full rotation for both angles to avoid overlaps.

📖 Explanation: The surface is generated by revolving $z = \ln u$ around the $z$-axis. The radius is $u$, and $f(u)=\ln u$ so f'(u)=1/u. The area element is $u\sqrt{1+(1/u)^2} \, du\,dv = u\sqrt{1+1/u^2} \, du\,dv$. Option A misses the radial factor $u$; B misidentifies the shape; C has incorrect derivative under the radical.

📖 Explanation: Both parametrizations are identical in form; the names of parameters are arbitrary. If $u$ and $s$ both represent radius and $v,t$ represent angle, they are the same mapping. The student's claim about swapped roles is incorrect because the structure $(radius \cdot \cos(angle), radius \cdot \sin(angle), radius^2)$ is preserved. Only domain restrictions could affect coverage.

📖 Explanation: Revolving $x = \cosh z$ about the $z$-axis gives $x = \cosh u, y = \sinh u$ style but with circular symmetry: $\mathbf{r}(u,v) = (\cosh u \cos v, \cosh u \sin v, u)$. This is a hyperboloid of one sheet because $x^2+y^2 = \cosh^2 u = 1+\sinh^2 u$ and $z=u$ gives $x^2+y^2 - z^2 = 1$. Option C names it incorrectly; D uses sinh which gives a different surface.

📖 Explanation: If $u$ is misinterpreted as height, one might incorrectly use the arc length formula for $z=u^2$ without the radial factor $u$. The correct area element is $u\sqrt{1+(2u)^2} = u\sqrt{1+4u^2}$. Option A shows the missing radial factor, a common error. Option C has incorrect derivative; D uses wrong derivative. Recognizing the role of parameters is essential.

📖 Explanation: Revolving $z = 2y+1$ about the $z$-axis sets the radial distance $u = y$, so $\mathbf{r}(u,v) = (u\cos v, u\sin v, 2u+1)$. The derivative with respect to $u$ is $(\cos v, \sin v, 2)$, with $v$ is $(-u\sin v, u\cos v, 0)$; cross product magnitude is $u\sqrt{5}$. Option A is correct but D is identical; the key is recognizing the radial factor $u$. Option B misses it; C swaps coordinates.

📖 Explanation: For a surface of revolution $(f(u)\cos v, f(u)\sin v, g(u))$, the Gaussian curvature is K = \frac{g'(u)g''(u)}{f(u)(1+[g'(u)]^2)^2}. Here g'(u)=3u^2, g''(u)=6u, so $K = \frac{18u^3}{u(1+9u^4)^2} > 0$ for $u>0$. Wait, check: actually g'(u)g''(u) = 18u^3, divided by $u$ gives $18u^2 >0$. So curvature is positive. Option C is correct; D incorrectly states negative. This tests deep understanding of differential geometry of surfaces of revolution.

📖 Explanation: The curve $y = \sqrt{x}$ means $x = y^2$. Revolving about the $y$-axis gives radius $r = y^2$ and height $y$. Parametrization $\mathbf{r}(u,v) = (u^2\cos v, u, u^2\sin v)$ has radius $u^2$ and height $u$, so it is correct. The area element for radius $f(u)=u^2$ and height $g(u)=u$ is |f(u)|\sqrt{[f'(u)]^2+[g'(u)]^2} = u^2\sqrt{(2u)^2+1} = u^2\sqrt{4u^2+1}. Wait, the student's integral has $u^2\sqrt{1+4u^2}$ which is actually correct. Let's re-evaluate: The student's integral matches the correct element. So the correct answer is A. But we must check: The derivative of $u^2$ is $2u$, so f'^2+g'^2 = 4u^2+1. So $u^2\sqrt{4u^2+1}$ is correct. So the student is correct. Option A is true.

📖 Explanation: The surface is generated by $z = e^u$ with radius $u$. Area element is $u\sqrt{1+(e^u)^2} = u\sqrt{1+e^{2u}}$. So the integral set-up is $\int_0^{2\pi}\int_0^1 u\sqrt{1+e^{2u}} \, du\,dv$. Option B misses the radial factor $u$; C uses wrong derivative; D has $e^u$ instead of $e^{2u}$. Proper computation requires chain rule and Jacobian.

📖 Explanation: The surface is generated by revolving $z = \sin u$ where $u$ is radial distance. Since $\sin u$ ranges from 0 to 1 for $u \in [0,\pi]$, but actually $\sin u$ is non-negative on $[0,\pi]$, so $z \in [0,1]$. However, the surface looks like a series of concentric sinusoidal ripples, not a cone or helicoid. Option A correctly states $z \in [0,1]$ but the description 'sine-wave surface' is vague; B says $z \in [-1,1]$ which is wrong. Actually correct is A with proper range. But careful: $\sin u$ for $u \in [0,\pi]$ gives values from 0 to 1, so range is [0,1]. So A is correct. The graph is a wavy bowl with no negative z. So answer A.

📖 Explanation: The cone $z = \sqrt{x^2+y^2}$ is parametrized by radius $u$ and angle $v$. Increasing the angle range from $[0,2\pi]$ to $[0,4\pi]$ means each point (except the apex) is covered twice because $\cos(v+2\pi)=\cos v$ and $\sin(v+2\pi)=\sin v$. So (II) covers the surface twice. The height is same since $u \in [0,1]$ in both. Option A is correct; B says four times which is wrong; C incorrectly changes height; D ignores periodicity.