Line Integrals: Scalar Fields - Multivariable Calculus

1. Let $C$ be a smooth curve of length $L>0$ in the plane, and let $f(x,y)$ be a continuous function defined on $C$. Which of the following statements about the scalar line integral $\int_C f \, ds$ is NOT always true?
   Answer: If $C$ is a closed curve, then $\int_C f \, ds = 0$. Let's analyze each statement: A) This is a standard property of integrals. The integral is bounded by the minimum and maximum values of the integrand multiplied by the measure of the domain (in this case, arc length $L$). This statement is always true. B) Scalar line integrals are defined in terms of arc length $ds$, which is inherently positive regardless of the direction of traversal. Therefore, the integral is independent of the orientation of the curve. This statement is always true. C) This is not always true. A simple counterexample is integrating a positive constant function over a closed curve. For example, let $C$ be the unit circle and $f(x,y)=1$. Then $\int_C 1 \, ds$ is the arc length of the unit circle, which is $2\pi$, not 0. This property (integral over a closed path is zero) is characteristic of conservative vector fields, not scalar line integrals in general. D) This is the linearity property of integrals, which holds for scalar line integrals. This statement is always true. Therefore, the statement that is not always true is C.
2. Let $C$ be the boundary of the rectangle with vertices $(0,0)$, $(a,0)$, $(a,b)$, and $(0,b)$, where $a > 0$ and $b > 0$. What is the value of the line integral $\int_C (x+y) \, ds$?
   Answer: $(a+b)^2$ We can split the integral into four parts, one for each side of the rectangle. $C_1$: from $(0,0)$ to $(a,0)$. Parameterize as $\mathbf{r}(t)=\langle t,0 \rangle$ for $0 \le t \le a$. $ds=dt$. $\int_{C_1} (x+y)ds = \int_0^a (t+0)dt = \frac{a^2}{2}$. $C_2$: from $(a,0)$ to $(a,b)$. Parameterize as $\mathbf{r}(t)=\langle a,t \rangle$ for $0 \le t \le b$. $ds=dt$. $\int_{C_2} (x+y)ds = \int_0^b (a+t)dt = ab + \frac{b^2}{2}$. $C_3$: from $(a,b)$ to $(0,b)$. Parameterize as $\mathbf{r}(t)=\langle a-t,b \rangle$ for $0 \le t \le a$. $ds=dt$. $\int_{C_3} (x+y)ds = \int_0^a (a-t+b)dt = [at-\frac{t^2}{2}+bt]_0^a = a^2 - \frac{a^2}{2} + ab = \frac{a^2}{2} + ab$. $C_4$: from $(0,b)$ to $(0,0)$. Parameterize as $\mathbf{r}(t)=\langle 0,b-t \rangle$ for $0 \le t \le b$. $ds=dt$. $\int_{C_4} (x+y)ds = \int_0^b (0+b-t)dt = [bt-\frac{t^2}{2}]_0^b = b^2 - \frac{b^2}{2} = \frac{b^2}{2}$. Summing the four parts: $\frac{a^2}{2} + (ab + \frac{b^2}{2}) + (\frac{a^2}{2} + ab) + \frac{b^2}{2} = a^2 + b^2 + 2ab = (a+b)^2$. Alternatively, we can compute $\int_C x \, ds$ and $\int_C y \, ds$ separately. $\int_C x \, ds = \frac{a^2}{2} + \int_0^b a \, dt + \int_0^a (a-t) \, dt + 0 = \frac{a^2}{2} + ab + \frac{a^2}{2} = a^2+ab = a(a+b)$. By symmetry (swapping roles of $a,b$ and $x,y$), $\int_C y \, ds = b(a+b)$. The sum is $a(a+b)+b(a+b) = (a+b)^2$. Distractors B and C represent these partial results. Distractor A is the result of evaluating the vector line integral $\oint_C \langle x,y \rangle \cdot d\mathbf{r}$, which is 0 by Green's Theorem, a common point of confusion.
3. A thin wire is bent in the shape of a quarter-circle of radius $R$ in the first quadrant, with endpoints at $(R,0)$ and $(0,R)$. The wire has a constant linear mass density. What is the $x$-coordinate of the center of mass of the wire?
   Answer: $\frac{2R}{\pi}$ The $x$-coordinate of the center of mass, $\bar{x}$, of a wire with constant density is given by $\bar{x} = \frac{\int_C x \, ds}{\int_C 1 \, ds}$. The denominator is the total length of the wire, $L$. The curve is a quarter-circle of radius $R$, so its length is $L = \frac{1}{4}(2\pi R) = \frac{\pi R}{2}$. To compute the numerator, we parameterize the quarter-circle $C$ as $\mathbf{r}(t) = \langle R\cos(t), R\sin(t) \rangle$ for $t \in [0, \pi/2]$. The arc length element is $ds = \|\mathbf{r}'(t)\| dt = \|\langle -R\sin(t), R\cos(t) \rangle\| dt = R \, dt$. The integral in the numerator is: $\int_C x \, ds = \int_0^{\pi/2} (R\cos(t)) (R \, dt) = R^2 \int_0^{\pi/2} \cos(t) \, dt = R^2 [\sin(t)]_0^{\pi/2} = R^2 (\sin(\pi/2) - \sin(0)) = R^2(1-0) = R^2.$ Now, we can find the $x$-coordinate of the center of mass: $\bar{x} = \frac{R^2}{L} = \frac{R^2}{\pi R / 2} = \frac{2R}{\pi}.$ Distractor A, $R/2$, is an intuitive but incorrect guess based on the range of x-values. Distractor C, $\frac{4R}{3\pi}$, is the $x$-coordinate of the center of mass of a solid quarter-circular plate (a lamina), which is a common confusion. Distractor D, $\frac{\pi R}{4}$, is half the arc length of the wire.
4. Let $C_1$ be the line segment from $(0,0)$ to $(1,1)$ and $C_2$ be the arc of the parabola $y=x^2$ from $(0,0)$ to $(1,1)$. Let $I_1 = \int_{C_1} x \, ds$ and $I_2 = \int_{C_2} x \, ds$. Which of the following statements is true?
   Answer: $I_1 < I_2$ We need to compute both integrals and compare them. For $I_1$: Parameterize $C_1$ by $\mathbf{r}(t) = \langle t, t \rangle$ for $0 \le t \le 1$. Then $\mathbf{r}'(t) = \langle 1, 1 \rangle$, so $ds = \|\mathbf{r}'(t)\| dt = \sqrt{1^2+1^2} \, dt = \sqrt{2} \, dt$. The function is $f(x,y)=x$, so $f(\mathbf{r}(t)) = t$. $I_1 = \int_0^1 t \sqrt{2} \, dt = \sqrt{2} [\frac{t^2}{2}]_0^1 = \frac{\sqrt{2}}{2}$. For $I_2$: Parameterize $C_2$ by $\mathbf{r}(t) = \langle t, t^2 \rangle$ for $0 \le t \le 1$. Then $\mathbf{r}'(t) = \langle 1, 2t \rangle$, so $ds = \|\mathbf{r}'(t)\| dt = \sqrt{1^2+(2t)^2} \, dt = \sqrt{1+4t^2} \, dt$. The function is $f(x,y)=x$, so $f(\mathbf{r}(t)) = t$. $I_2 = \int_0^1 t \sqrt{1+4t^2} \, dt$. We use u-substitution with $u = 1+4t^2$, so $du = 8t \, dt$. The limits change from $t=0,1$ to $u=1,5$. $I_2 = \int_1^5 \sqrt{u} \frac{du}{8} = \frac{1}{8} [\frac{2}{3}u^{3/2}]_1^5 = \frac{1}{12} (5^{3/2} - 1^{3/2}) = \frac{5\sqrt{5}-1}{12}$. Now we compare $I_1 = \frac{\sqrt{2}}{2}$ and $I_2 = \frac{5\sqrt{5}-1}{12}$. We can approximate: $I_1 \approx \frac{1.414}{2} = 0.707$. For $I_2$, $\sqrt{5} \approx 2.236$, so $5\sqrt{5} \approx 11.18$. Then $I_2 \approx \frac{11.18-1}{12} = \frac{10.18}{12} \approx 0.848$. Since $0.707 < 0.848$, we have $I_1 < I_2$. Distractor C is incorrect; it implies path independence, which is a property of line integrals of conservative vector fields, not scalar fields. Distractor D is incorrect as scalar line integrals are independent of parameterization.
5. The average value of a function $f$ over a curve $C$ is given by the formula $\frac{1}{L} \int_C f \, ds$, where $L$ is the arc length of $C$. Which expression represents the average value of $f(x,y) = x$ over the curve $C$ given by the parabola $y=x^2$ from $(0,0)$ to $(1,1)$?
   Answer: $\frac{\int_0^1 t \sqrt{1+4t^2} \, dt}{\int_0^1 \sqrt{1+4t^2} \, dt}$ The average value is $\frac{1}{L} \int_C f \, ds$. The arc length $L$ is itself a line integral: $L = \int_C 1 \, ds$. So the average value is $\frac{\int_C f \, ds}{\int_C 1 \, ds}$. We parameterize the curve $C$ by $\mathbf{r}(t) = \langle t, t^2 \rangle$ for $0 \le t \le 1$. The derivative is $\mathbf{r}'(t) = \langle 1, 2t \rangle$, so the arc length element is $ds = \|\mathbf{r}'(t)\| dt = \sqrt{1^2+(2t)^2} \, dt = \sqrt{1+4t^2} \, dt$. For the numerator, we integrate $f(x,y)=x$. In terms of the parameter, $f(\mathbf{r}(t)) = t$. So, $\int_C x \, ds = \int_0^1 t \sqrt{1+4t^2} \, dt$. For the denominator, we integrate $1$ to find the arc length $L$. So, $L = \int_C 1 \, ds = \int_0^1 1 \cdot \sqrt{1+4t^2} \, dt$. Combining these, the average value is $\frac{\int_0^1 t \sqrt{1+4t^2} \, dt}{\int_0^1 \sqrt{1+4t^2} \, dt}$. Distractor B is the expression for the average value if one incorrectly assumes $ds=dt$, ignoring the geometry of the curve. Distractor C is just the numerator, the integral of the function, not the average value. Distractor D is the average value of the function $g(x,y)=y$ over the curve, not $f(x,y)=x$.
6. Let $C$ be the square with vertices $(1,0)$, $(0,1)$, $(-1,0)$, and $(0,-1)$. What is the value of $\int_C (x^3 + \sin y) \, ds$?
   Answer: $0$ The integral can be split by linearity: $\int_C (x^3 + \sin y) \, ds = \int_C x^3 \, ds + \int_C \sin y \, ds$. We can evaluate this integral using symmetry arguments. The curve $C$ (a square centered at the origin) is symmetric with respect to both the x-axis and the y-axis. Consider the integral $\int_C x^3 \, ds$. The function $f(x,y)=x^3$ is an odd function of $x$. For any point $(x,y)$ on the square, the point $(-x,y)$ is also on the square (if $y>0$, reflection across y-axis; if $y<0$, reflection across y-axis). The value of $ds$ is the same for corresponding symmetric segments. Since $f(-x,y) = (-x)^3 = -x^3 = -f(x,y)$, the contributions to the integral from the right half of the square (where $x>0$) and the left half (where $x<0$) will cancel each other out. Thus, $\int_C x^3 \, ds = 0$. Similarly, consider $\int_C \sin y \, ds$. The function $g(x,y)=\sin y$ is an odd function of $y$. The square is symmetric with respect to the x-axis. For any point $(x,y)$, the point $(x,-y)$ is also on the curve. Since $g(x,-y) = \sin(-y) = -\sin y = -g(x,y)$, the contributions from the upper half ($y>0$) and lower half ($y<0$) cancel out. Thus, $\int_C \sin y \, ds = 0$. The total integral is $0+0=0$. Distractor C, $4\sqrt{2}$, is the length of the square (each side has length $\sqrt{2}$). This would be the answer for $\int_C 1 \, ds$. Distractors B and D tempt students who do not see the symmetry argument into thinking a complex calculation is required.
7. Let $C$ be the unit circle $x^2+y^2=1$. It is given that $\int_C (x^2+y^2) \, ds = 2\pi$. Let $C_R$ be the circle $x^2+y^2=R^2$ with $R>0$. What is the value of the line integral $\int_{C_R} (x^2+y^2) \, ds$?
   Answer: $2\pi R^3$ We can solve this by direct computation. Parameterize the circle $C_R$ by $\mathbf{r}(t) = \langle R\cos(t), R\sin(t) \rangle$ for $0 \le t \le 2\pi$. The derivative is $\mathbf{r}'(t) = \langle -R\sin(t), R\cos(t) \rangle$. The arc length element is $ds = \|\mathbf{r}'(t)\| dt = \sqrt{(-R\sin t)^2 + (R\cos t)^2} \, dt = \sqrt{R^2(\sin^2 t + \cos^2 t)} \, dt = R \, dt$. The function is $f(x,y) = x^2+y^2$. On the curve $C_R$, the value of the function is constant: $f(R\cos t, R\sin t) = (R\cos t)^2 + (R\sin t)^2 = R^2$. The integral is: $\int_{C_R} (x^2+y^2) \, ds = \int_0^{2\pi} (R^2) \cdot (R \, dt) = R^3 \int_0^{2\pi} dt = R^3 [t]_0^{2\pi} = 2\pi R^3.$ Alternatively, one can use a scaling argument. Let $f(x,y)=x^2+y^2$ be a homogeneous function of degree $k=2$. If a curve $C$ is scaled by a factor $R$ to become $C_R$, the line integral scales as $\int_{C_R} f \, ds = R^{k+1} \int_C f \, ds$. Here $k=2$, so the integral scales by $R^{2+1}=R^3$. Given $\int_C (x^2+y^2) \, ds = 2\pi$, the new integral is $R^3(2\pi)$. Distractor A, $2\pi R$, is the length of the circle $C_R$, which equals $\int_{C_R} 1 \, ds$. Distractor B, $2\pi R^2$, results from correctly evaluating the function as $R^2$ on the circle but forgetting that $ds$ also scales with $R$.
8. Let $C$ be the cardioid given in polar coordinates by $r = 1 + \cos(\theta)$ for $0 \le \theta \le 2\pi$. Evaluate the line integral $\int_C \sqrt{x^2+y^2} \, ds$.
   Answer: $\frac{32}{3}$ The integrand is $\sqrt{x^2+y^2}$, which in polar coordinates is simply $r$. So we need to compute $\int_C r \, ds$. The arc length element in polar coordinates is $ds = \sqrt{r^2 + (dr/d\theta)^2} \, d\theta$. Here, $r = 1+\cos(\theta)$, so $dr/d\theta = -\sin(\theta)$. $ds = \sqrt{(1+\cos\theta)^2 + (-\sin\theta)^2} \, d\theta = \sqrt{1+2\cos\theta+\cos^2\theta+\sin^2\theta} \, d\theta = \sqrt{2+2\cos\theta} \, d\theta$. Using the half-angle identity $1+\cos\theta = 2\cos^2(\theta/2)$, we get $ds = \sqrt{4\cos^2(\theta/2)} \, d\theta = 2|\cos(\theta/2)| \, d\theta$. The integral becomes $\int_0^{2\pi} (1+\cos\theta) \cdot 2|\cos(\theta/2)| \, d\theta$. We must split the integral because of the absolute value. For $0 \le \theta \le \pi$, $\cos(\theta/2) \ge 0$. For $\pi < \theta \le 2\pi$, $\cos(\theta/2) < 0$. Again using $1+\cos\theta = 2\cos^2(\theta/2)$, the integrand is $(2\cos^2(\theta/2)) \cdot 2|\cos(\theta/2)| = 4|\cos^3(\theta/2)|$. Integral $= \int_0^{\pi} 4\cos^3(\theta/2) \, d\theta + \int_{\pi}^{2\pi} -4\cos^3(\theta/2) \, d\theta$. Let's evaluate the first part: $4\int_0^{\pi} \cos^3(\theta/2) \, d\theta = 4\int_0^{\pi} (1-\sin^2(\theta/2))\cos(\theta/2) \, d\theta$. Let $u=\sin(\theta/2)$, $du = \frac{1}{2}\cos(\theta/2)d\theta$. Limits become $\sin(0)=0$ to $\sin(\pi/2)=1$. The integral is $4 \int_0^1 (1-u^2) 2du = 8[u-u^3/3]_0^1 = 8(1-1/3) = 16/3$. By symmetry of the cardioid, the integral from $\pi$ to $2\pi$ will also be $16/3$. So the total is $16/3+16/3 = 32/3$. Distractor A, $0$, arises from ignoring the absolute value and integrating $\cos^3(\theta/2)$ over a full period, which gives zero. Distractor B, $16/3$, is the result of integrating over only half the curve. Distractor D, $2\pi$, is the result of computing $\int_0^{2\pi} (1+\cos\theta)d\theta$, which ignores the $ds$ factor.
9. Consider the curve $C$ that consists of the semicircle $x^2 + y^2 = 1$ with $y \geq 0$, traversed counterclockwise, followed by the line segment from $(-1, 0)$ to $(1, 0)$. If $f(x,y) = xy + 1$, what is $\int_C f(x,y) \, ds$?
   Answer: $\pi + 2$ We split the integral into two parts. For $C_1$ (semicircle): parameterize as $\vec{r}_1(t) = (\cos t, \sin t)$ for $0 \leq t \leq \pi$. Then $f(\cos t, \sin t) = \cos t \sin t + 1 = \frac{\sin 2t}{2} + 1$ and $\|\vec{r}_1'(t)\| = \sqrt{\sin^2 t + \cos^2 t} = 1$. So $\int_{C_1} f(x,y) \, ds = \int_0^\pi (\frac{\sin 2t}{2} + 1) \, dt = [-\frac{\cos 2t}{4} + t]_0^\pi = [-\frac{1}{4} + \pi] - [-\frac{1}{4} + 0] = \pi$. For $C_2$ (line segment from $(-1,0)$ to $(1,0)$): parameterize as $\vec{r}_2(t) = (-1 + 2t, 0)$ for $0 \leq t \leq 1$. Then $f(-1 + 2t, 0) = (-1 + 2t) \cdot 0 + 1 = 1$ and $\|\vec{r}_2'(t)\| = 2$. So $\int_{C_2} f(x,y) \, ds = \int_0^1 1 \cdot 2 \, dt = 2$. Therefore, $\int_C f(x,y) \, ds = \pi + 2$.
10. Let $C$ be the upper semicircle of radius 2 centered at the origin, traversed from $(2,0)$ to $(-2,0)$. The value of the line integral $\int_C (x^2+y) \, ds$ is $K$. What is the value of the line integral $\int_{C'} (x^2+y) \, ds$, where $C'$ is the lower semicircle of radius 2, also traversed from $(2,0)$ to $(-2,0)$?
    Answer: $K-16$ Let's calculate $K$. The upper semicircle $C$ can be parameterized by $\mathbf{r}(t) = \langle 2\cos(t), 2\sin(t) \rangle$ for $t$ from $0$ to $\pi$. Then $\mathbf{r}'(t) = \langle -2\sin(t), 2\cos(t) \rangle$, so $ds = \|\mathbf{r}'(t)\| dt = \sqrt{4\sin^2(t) + 4\cos^2(t)} \, dt = 2 \, dt$. The integrand is $f(x,y) = x^2+y = (2\cos t)^2 + 2\sin t = 4\cos^2 t + 2\sin t$. So, $K = \int_0^{\pi} (4\cos^2 t + 2\sin t) (2 \, dt) = 8\int_0^{\pi} \cos^2 t \, dt + 4\int_0^{\pi} \sin t \, dt.$ Using $\cos^2 t = \frac{1+\cos(2t)}{2}$, the first integral is $8[\frac{t}{2} + \frac{\sin(2t)}{4}]_0^{\pi} = 8(\frac{\pi}{2}) = 4\pi$. The second integral is $4[-\cos t]_0^{\pi} = 4(1 - (-1)) = 8$. Thus, $K = 4\pi + 8$. Now, let's calculate the integral over the lower semicircle $C'$. We can parameterize $C'$ as $\mathbf{g}(t) = \langle 2\cos(t), -2\sin(t) \rangle$ for $t$ from $0$ to $\pi$. Then $\mathbf{g}'(t) = \langle -2\sin(t), -2\cos(t) \rangle$, and $ds = \|\mathbf{g}'(t)\| dt = 2 \, dt$. The integrand is $f(x,y) = x^2+y = (2\cos t)^2 + (-2\sin t) = 4\cos^2 t - 2\sin t$. So, $\int_{C'} (x^2+y)\,ds = \int_0^{\pi} (4\cos^2 t - 2\sin t) (2 \, dt) = 8\int_0^{\pi} \cos^2 t \, dt - 4\int_0^{\pi} \sin t \, dt.$ This evaluates to $4\pi - 8$. Comparing this to $K = 4\pi+8$, we see that the new value is $(4\pi+8) - 16 = K-16$. Distractor A, $K$, would be correct if the integrand were symmetric with respect to the x-axis (e.g., if it were just $x^2$). Distractor B, $-K$, is a common mistake confusing scalar line integrals with vector line integrals, where reversing path direction can negate the value. Distractor C, $K-8$, is a possible calculation error.
11. Let $C$ be the arc of the parabola $y = x^2$ from $(0,0)$ to $(2,4)$. Which of the following integrals represents the value of $\int_C e^{xy} \, ds$?
    Answer: $\int_0^2 e^{t^3} \sqrt{1+4t^2} \, dt$ To set up the line integral, we first need to parameterize the curve $C$. Since $y=x^2$, we can use $x$ as the parameter, say $x=t$. Then $y=t^2$. The path starts at $(0,0)$ (so $t=0$) and ends at $(2,4)$ (so $t=2$). The parameterization is $\mathbf{r}(t) = \langle t, t^2 \rangle$ for $t \in [0,2]$. Next, we find the arc length element $ds$. The derivative is $\mathbf{r}'(t) = \langle 1, 2t \rangle$. Its magnitude is $\|\mathbf{r}'(t)\| = \sqrt{1^2 + (2t)^2} = \sqrt{1+4t^2}$. So, $ds = \sqrt{1+4t^2} \, dt$. Finally, we express the function $f(x,y) = e^{xy}$ in terms of the parameter $t$: $f(\mathbf{r}(t)) = e^{(t)(t^2)} = e^{t^3}$. Combining these pieces, the line integral is $\int_C f \, ds = \int_0^2 f(\mathbf{r}(t)) \|\mathbf{r}'(t)\| \, dt = \int_0^2 e^{t^3} \sqrt{1+4t^2} \, dt$. Distractor B incorrectly assumes $ds = dt$, omitting the arc length factor $\|\mathbf{r}'(t)\|$. Distractor C incorrectly substitutes $y=t$ into the function, evaluating $e^{xt}$ as $e^{t^2}$ instead of $e^{x(t)y(t)} = e^{t \cdot t^2} = e^{t^3}$. Distractor D results from an incorrect attempt to parameterize with respect to $y$, with errors in both the function evaluation and the arc length element calculation.
12. Let $C$ be the path consisting of two line segments: from $(1,0)$ to $(0,1)$ and then from $(0,1)$ to $(-1,0)$. Compute the line integral $\int_C x^2 \, ds$.
    Answer: $\frac{2\sqrt{2}}{3}$ The path $C$ is composed of two segments, $C_1$ from $(1,0)$ to $(0,1)$ and $C_2$ from $(0,1)$ to $(-1,0)$. We compute the integral over each segment and add the results. For $C_1$, we can parameterize the path as $\mathbf{r}_1(t) = \langle 1-t, t \rangle$ for $t \in [0,1]$. Then $\mathbf{r}_1'(t) = \langle -1, 1 \rangle$, and the arc length element is $ds = \|\mathbf{r}_1'(t)\| dt = \sqrt{(-1)^2 + 1^2} \, dt = \sqrt{2} \, dt$. The function is $f(x,y)=x^2$, so on the path $f(\mathbf{r}_1(t)) = (1-t)^2$. The integral over $C_1$ is: $\int_{C_1} x^2 \, ds = \int_0^1 (1-t)^2 \sqrt{2} \, dt = \sqrt{2} \left[ -\frac{(1-t)^3}{3} \right]_0^1 = \sqrt{2} \left(0 - \left(-\frac{1}{3}\right)\right) = \frac{\sqrt{2}}{3}.$ For $C_2$, we can parameterize the path as $\mathbf{r}_2(t) = \langle -t, 1-t \rangle$ for $t \in [0,1]$. Then $\mathbf{r}_2'(t) = \langle -1, -1 \rangle$, and $ds = \|\mathbf{r}_2'(t)\| dt = \sqrt{(-1)^2 + (-1)^2} \, dt = \sqrt{2} \, dt$. The function on the path is $f(\mathbf{r}_2(t)) = (-t)^2 = t^2$. The integral over $C_2$ is: $\int_{C_2} x^2 \, ds = \int_0^1 t^2 \sqrt{2} \, dt = \sqrt{2} \left[ \frac{t^3}{3} \right]_0^1 = \frac{\sqrt{2}}{3}.$ The total integral is the sum: $\int_C x^2 \, ds = \frac{\sqrt{2}}{3} + \frac{\sqrt{2}}{3} = \frac{2\sqrt{2}}{3}$. Distractor A is incorrect; the integrand is always non-negative. Distractor B is the result of forgetting the $\sqrt{2}$ factor from $ds$ in both parts. Distractor C is the result of computing the integral over only one of the two segments.
13. A wire is shaped like a helix parameterized by $\mathbf{r}(t) = \langle \cos(t), \sin(t), t \rangle$ for $0 \le t \le 2\pi$. The density of the wire at any point $(x,y,z)$ is given by $\rho(x,y,z) = z$. What is the total mass of the wire?
    Answer: $2\pi^2\sqrt{2}$ The mass of the wire is given by the line integral of its density function over the curve $C$, which is $M = \int_C \rho(x,y,z) \, ds$. First, we find the arc length element $ds$. The parameterization is $\mathbf{r}(t) = \langle \cos(t), \sin(t), t \rangle$. The derivative is $\mathbf{r}'(t) = \langle -\sin(t), \cos(t), 1 \rangle$. The magnitude of the derivative is $\|\mathbf{r}'(t)\| = \sqrt{(-\sin(t))^2 + (\cos(t))^2 + 1^2} = \sqrt{\sin^2(t) + \cos^2(t) + 1} = \sqrt{1+1} = \sqrt{2}$. So, $ds = \sqrt{2} \, dt$. The density function along the curve is $\rho(\mathbf{r}(t)) = z(t) = t$. Now we set up and evaluate the integral for the mass: $M = \int_0^{2\pi} t \cdot \sqrt{2} \, dt = \sqrt{2} \int_0^{2\pi} t \, dt = \sqrt{2} \left[ \frac{t^2}{2} \right]_0^{2\pi} = \sqrt{2} \left( \frac{(2\pi)^2}{2} - 0 \right) = \sqrt{2} \frac{4\pi^2}{2} = 2\pi^2\sqrt{2}$. Distractor A, $2\pi^2$, is the result of forgetting the $\sqrt{2}$ factor from the arc length element $ds$, effectively calculating $\int_C \rho \, dt$. Distractor B, $2\pi\sqrt{2}$, is the result of integrating a density of 1 instead of $z=t$, which would calculate the length of the wire multiplied by $\sqrt{2}$ for some reason. The length is $2\pi\sqrt{2}$. Distractor D, $0$, would be the result if the integral of an odd function were taken over a symmetric interval like $[-\pi, \pi]$, but the given interval is $[0, 2\pi]$.

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