Math 234 — Final Exam Review

Line integrals

Line Integral $L_1$

Short solution
The integral \[ L_1=\int_\cC 3ds = 3\int_\cC ds \] is three times the length of $\cC=AB$. The distance from $A$ to $B$ is $\sqrt{3^2+4^2}=5$, so \[ L_1 = 3\cdot 5 = 15. \]
Standard solution
Instead, you could also find a parametrization of the line segment and use $ds=\|\vx'(t)\|\,dt$. In this case that's more complicated, but here is how it goes: We have \[ \vx(t) = \va + t(\vb-\va) = \begin{pmatrix} 4t\\ 3-3t \end{pmatrix} \] so \[ \vx'(t) = \begin{pmatrix} 4\\ -3 \end{pmatrix}a \implies \|\vx'(t)\| = \sqrt{4^2+(-3)^2} = 5. \] Hence \[ \int_\cC 3\,ds = \int _0^1 3\,\|\vx'(t)\|\,dt = \int _0^1 3\cdot\,dt =15. \]

Line integral $L_2$

The vector \[ \vAB = \begin{pmatrix} 4\\-3\end{pmatrix} \] is a tangent vector to the line segment $AB$, but it is not a unit vector. To get the unit tangent divide by the length of $AB$: \[ \vT=\frac{\vAB}{\|\vAB\|} = \begin{pmatrix} 4/5\\-3/5\end{pmatrix}. \] It follows that \[ \vF\cdot\vT = \frac 45 \cdot 2 +(-\frac35)(-1) = \frac{11}5, \] and the work integral \[ L_2 = \int_\cC \vF\cdot\vT \,ds = \int_\cC \frac{11}{5} \,ds = \frac{11}5\cdot\text{(length of $\cC$}) = 11. \]

Line integral $L_3$

To get the upward normal from the tangent, rotate it: \[ \vT = \begin{pmatrix} 4/5\\-3/5\end{pmatrix} \implies \vN = \begin{pmatrix} 3/5\\4/5\end{pmatrix} \] Thus \[ \vF\cdot\vN = (\frac 35)(2) +(\frac45)(-1) = \frac25, \] and \[ L_3 = \int_\cC \vF\cdot\vN\, ds = \frac25\,\bigl(\text{length $\cC$}\bigr) =\frac25\cdot 5 = 2. \]

The flux and work done by $\vF$ across/along $AB$

The unit tangent vector $\vT$ is a vector of length one, in the same direction as the interval $AB$. Therefore we can compute \[ \vF\cdot\vT = \|\vF\| \, \|\vT\|\,\cos\theta =3\cdot1\cdot\cos\theta =3\cos\theta. \] Hence \[ \text{Work} = \int_{AB} \vF\cdot\vT\, ds = (3\cos\theta)(\text{length }AB) = 21\cos\theta. \] To compute the flux, we must find $\vN\cdot\vF$. The unit normal $\vN$ is perpendicular to $AB$, and if we choose the normal that points upward, then the angle between $\vN$ and $\vT$ is $\frac\pi2-\theta$. Therefore \[ \vN\cdot\vF = \|\vN\| \, \|\vF\| \, \cos\bigl(\frac\pi2-\theta\bigr) =1\cdot3\cdot\sin\theta = 3\sin\theta. \] The flux across $AB$ in the upward direction is \[ \text{Flux} = \int_{AB}\vF\cdot\vN\, ds = \int_{AB} 3\sin\theta\, ds =(3\sin\theta)(\text{length }AB) =21\sin\theta. \]

Match the integrals

Integrals $A$, $C$, and $D$ are the same, namely they represent the work done by the vector field $\vF$ along $\cC$. Integrals $B$ and $E$ are both equal to the flux of $\vF$ across $\cC$.

Green's Theorem

See the text for precise statements of Green's theorem. See also the line integral summary. We have \[ \int_\cC \vF\cdot\vT\, ds = \int_\cC P(x, y)dx + Q(x, y)dy =\iint_\cR \bigl\{\frac{\pd Q}{\pd x}-\frac{\pd P}{\pd y}\bigr\}\, dA \] if $\vT$ is the counterclockwise unit tangent, and \[ \int_\cC \vF\cdot\vN\, ds = \int_\cC -Q(x, y)dx + P(x, y)dy =\iint_\cR \bigl\{\frac{\pd P}{\pd x} + \frac{\pd Q}{\pd y}\bigr\}\, dA \] if $\vN$ is the outward unit normal.

Therefore, if we know that \[ \frac{\pd P}{\pd x} + \frac{\pd Q}{\pd y} = 0, \] then Green's theorem implies that \[ \int_\cC \vF\cdot\vN\, ds = 0. \] On the other hand, Green's theorem does not say anything about the work integral \[ \int_\cC \vF\cdot\vT\, ds. \]

Integral of a gradient

The important thing to remember here is “the fundamental theorem of line integrals” (theorem 6.1, p.150 in the text), which says that \[ \int_\cC \vec\nabla f \cdot \vT\, ds = f(B) - f(A) \] for any path $\cC$ and any function $f$, if $A$ and $B$ are the initial and final points of the curve $\cC$.

For the problem in question, we have $A=(0,0)$, and $B=(\pi,1)$, and therefore \[ \int_\cC \vec\nabla f \cdot \vT\, ds = f(\pi,1) - f(0,0). \] The fact that the path $\cC$ is the straight line segment from $A$ to $B$ does not matter in this problem. The answer would have been the same for any other path.

The function is $f(x,y) = xy\cos(x)$, so \[ \int_\cC \vec\nabla f \cdot \vT\, ds = \pi\cdot 1\cdot \cos\pi - 0\cdot0\cdot\cos 0 = -\pi. \] Different solution: you could also calculate \[ \vec\nabla f = \begin{pmatrix} y\cos x-xy\sin y \\ x\cos x \end{pmatrix} \] parametrize the line segment $\cC$, and compute the (ugly) integral. This is in principle correct, and if you don't make mistakes, you should get the same answer. It was not the intended solution.

True/False questions about line integrals

The following figure shows some level sets of a function $f(x,y)$ (in grey), some points $K$, $L$, $M$, $N$, and $O$, and four paths $\cC_1$, … , $\cC_4$.

line integral of gradient picture

  1. $\int_{\cC_2} f ds = 0.2$ ? NOT TRUE, meaning, not necessarily true: all we know is that $f(x,y)$ lies between 0.1 and 0.5. No scale is included with the drawing, so we can’t tell if the length of the curve is 1, 100, or 10-34.
  2. $\int_{\cC_2} \vec\nabla f\cdot d\vec x = 0.2$ ? TRUE. The Fundamental Theorem for Line integrals applies and tells us that \[ \int_{\cC_2} \vec\nabla f\cdot d\vec x = f(N)-f(M) = 0.3-0.1=0.2 \]
  3. $\int_{\cC_2} f ds \gt 0$ ? TRUE. The Fundamental Theorem does not apply, so we cannot directly compute the integral. However, the function we are integrating is $f(x,y)$, and from the drawing of the curve $\cC_2$ and the level sets of $f$ we can see that $f(x,y)\ge 0.1$ on $\cC_2$. In particular, $f(x,y)>0$ on the curve $\cC_2$. This implies \[ \int_{\cC_2} f\,ds >0. \] “Why?” you ask. Well, to compute the integral you Since $f(x,y)>0$, and since $ds$ is a length, all the terms $f(x,y)ds$ are positive, and therefore the integral is also positive.
  4. $\int_{\cC_2} \vec\nabla f\cdot d\vec x \gt 0$ ? TRUE. We just computed the integral in question b above. It’s $0.2$, so it is certainly positive.
  5. $\int_{\cC_2} f ds = 0$ ? FALSE, because we have already shown that the integral is positive (c, above).
  6. $\int_{\cC_1} \vec\nabla f\cdot d\vec x \gt 0$ ? FALSE. The Fundamental Theorem applies, and we find that \[ \int_{\cC_1} \vec\nabla f\cdot d\vec x = f(M)-f(L) = 0.1-0.1=0. \]
  7. $\int_{-\cC_1} f ds \lt 0$ ? FALSE. For line integrals of the form $\int fds$ changing the direction (orientation) of the curve does not change the integral. So \[ \int_{-\cC_1} f\,ds = \int _{\cC_1}f\, ds. \] The reasoning in question c (above) tells us that the line integral $\int_\cC f(x,y)ds$ is positive whenever the function $f(x,y)$ is positive. Since $f(x,y)\ge 0.1 \gt 0$ on $\cC_1$, we have $\int_{\cC_1}f\,ds\gt0$, and thus also $\int_{-\cC_1}f\,ds\gt0$.
  8. $\int_{- \cC_1} \vec\nabla f\cdot d\vec x \gt 0$ ? FALSE. For “work” line integrals changing the orientation of the curve changes the sign of the integral. One way to think about this is to say that changing the direction of $\cC$ changes the direction of the vector $d\vx$, which then changes the sign of the integral. So here we have \[ \int_{-\cC_1}\vec\nabla f\cdot d\vx = -\int_{\cC_1}\vec\nabla f\cdot d\vx. \] In this case it turns out not to make any difference, because both integrals turn out to be zero. We find that by using the Fundamental Theorem, with result \[ \int_{-\cC_1}\vec\nabla f\cdot d\vx = f(L) - f(M) = 0.1-0.1=0. \] Note the order of $L$ and $M$ here: $L$ is the end point of $-\cC_1$, while $M$ is its initial point.
    9. line integral of gradient picture
  9. $\int_{\cC_1+\cC_2} \vec\nabla f\cdot d\vec x = \int_{\cC_2} \vec\nabla f\cdot d\vec x$ ? TRUE. By definition \[ \int_{\cC_1+\cC_2} \vec\nabla f\cdot d\vec x = \int_{\cC_1} \vec\nabla f\cdot d\vec x +\int_{\cC_2} \vec\nabla f\cdot d\vec x \] and the integral over $\cC_1$ is zero (see the previous question.)
  10. $\int_{\cC_1+\cC_2} \vec\nabla f\cdot d\vec x = \int_{\cC_4} \vec\nabla f\cdot d\vec x$ ? TRUE. The Fundamental Theorem implies \[ \int_{\cC_1+\cC_2} \vec\nabla f\cdot d\vec x = f(N)-f(L) \] and \[ \int_{\cC_4} \vec\nabla f\cdot d\vec x = f(O) - f(K). \] Since $N$ and $O$ are on the same level set of $f$ we have $f(N)=f(O) = 0.3$. Also, we have $f(K)=f(L)=0.1$. Therefore the two integrals are the same.
  11. $\int_{\cC_3} \vec\nabla f\cdot d\vec x \gt 0$ ? FALSE. $\cC_3$ is a closed curve. The Fundamental Theorem therefore implies that $\int_{\cC_3} \vec\nabla f\cdot d\vx = 0$, no matter which function $f$ is.