# Winter 2006 Final

The exam itself is available on WebCT and on Docuum. Didn't bother to type up the questions.

## 1Question 1¶

The equations

\begin{align*} xyuv + 2 = 0 \\ x^2+2y^2+3u^2+4v^2-19 = 0\end{align*}

define $u$, $v$ as functions of $x$, $y$ in the neighbourhood of the point $x=1$, $y=-1$, $u=2$, $v=1$.

Find

$$\begin{pmatrix} \frac{\partial u}{\partial x} & \frac{\partial v}{\partial x} \\ \frac{\partial u}{\partial y} & \frac{\partial v}{\partial y} \end{pmatrix}$$

at the point.

### 1.1Solution¶

First, we rewrite the system of equations as implicit functions of $x, y, u$ and $v$:

\begin{align*}F(x, y, u, v) & = xyuv - 2 = 0 \\ G(x, y, u, v) & = x^2+2y^2+3u^2+4v^2-19 = 0\end{align*}

The Jacobian $J$, at the point $x=1, y = -1, u = 2, v = 1$, is:

$$J = \frac{\partial(F, G)}{\partial (x, y)} = \det \begin{pmatrix} F_u & F_ v \\ G_u & G_v \end{pmatrix} = \det \begin{pmatrix} xyv & xyu \\ 6u & 8v \end{pmatrix} = \det\begin{pmatrix} -1 & -2 \\ 12 & 8 \end{pmatrix} = -8+24 = 16$$

(Recall that the "new variables", i.e. the dependent variables, are on the bottom.)

Now we find the partial derivatives of $u$ and $v$ with respect to $x$ and $y$:

\begin{align*} \frac{\partial u}{\partial x} & = -\frac{1}{J} \frac{\partial (F, G)}{\partial (x, v)} = -\frac{1}{16} \det \begin{pmatrix} yuv & xyu \\ 2x & 8v \end{pmatrix} = -\frac{1}{16} \det\begin{pmatrix} -2 & -2 \\ 2 & 8 \end{pmatrix} = -\frac{1}{16} (-16+4) = -\frac{-12}{16} = \frac{3}{4} \\ \frac{\partial u}{\partial y} & = -\frac{1}{J} \frac{\partial (F, G)}{\partial (y, v)} = -\frac{1}{16} \det\begin{pmatrix} xuv & xyu \\ 4y & 8v \end{pmatrix} = -\frac{1}{16} \det\begin{pmatrix} 2 & -2 \\ -4 & 8 \end{pmatrix} = -\frac{1}{16} (16 - 8) = -\frac{8}{16} = -\frac{1}{2} \\ \frac{\partial v}{\partial x} & = -\frac{1}{J} \frac{\partial (F, G)}{\partial (u, x)} = -\frac{1}{16} \det\begin{pmatrix} xyv & yvu \\ 6u & 2x \end{pmatrix} = -\frac{1}{16} \det\begin{pmatrix} -1 & -2 \\ 12 & 2 \end{pmatrix} = -\frac{1}{16} (-2 + 24) = -\frac{22}{16} = -\frac{3}{8} \\ \frac{\partial v}{\partial y} & = -\frac{1}{J} \frac{\partial (F, G)}{\partial (u, y)} = -\frac{1}{16} \det\begin{pmatrix} xyv & xuv \\ 6u & 4y \end{pmatrix} = -\frac{1}{16} \det\begin{pmatrix} -1 & 2 \\ 12 & -4 \end{pmatrix} = -\frac{1}{16} (4 - 24) = -\frac{-20}{16} = \frac{5}{4} \end{align*}

Consequently, the matrix we're looking for is

$$\begin{pmatrix} \frac{3}{4} & -\frac{3}{8} \\ -\frac{1}{2} & \frac{5}{4} \end{pmatrix}$$

Maybe

## 2Question 2¶

Lagrange multipliers aren't in the syllabus this year I believe

## 3Question 3¶

Let $R$ be the region in the first quadrant of the $xy$-plane bounded by the lines $y=x$ and $y=2x$ and by the hyperbolas $xy=1$ and $xy=3$.

Evaluate $\displaystyle \iint_R x^3y\,dA$.

### 3.1Solution¶

We can use a change of variables. Let $u = \frac{y}{x}$ and $v = xy$. We first find the Jacobian, as follows (again note that the "new variables" are on the bottom, and that we take 1 over that Jacobian to get one which we can compute):

$$\frac{\partial (x, y)}{\partial (u, v)} = 1 / \frac{\partial (u, v)}{\partial (x, y)} = 1 /\det \begin{pmatrix} u_x & u_y \\ v_x & v_y \end{pmatrix} = 1 / \det \begin{pmatrix} -\frac{y}{x^2} & \frac{1}{x} \\ y & x \end{pmatrix} = 1 / \left ( -\frac{2y}{x} \right ) = -\frac{x}{2y}$$

Note that we take the absolute value of the Jacobian in this case (though not in the implicit function question above), because it's a scaling factor. So the value that we'll need is $\displaystyle \frac{x}{2y}$.

Now, we multiply the integrand by the Jacobian, and express that in terms of $u$ and $v$:

$$x^3y \cdot \frac{x}{2y} = \frac{1}{2} x^4 = \frac{1}{2} \left ( \frac{v}{u} \right )^2$$

(Note that we could have first expressed the Jacobian in terms of $u$ and $v$ and then multiplied that by the integrand - the order is irrelevant.)

The limits are $u = 1$ to $u = 2$ and $v = 1$ to $v = 3$. The integral therefore becomes

$$\frac{1}{2} \int_1^2 \int_1^3 \frac{v^2}{u^2} \,dv\,du = \frac{1}{2} \int_1^2 \left [ \frac{v^3}{3u^2} \right ]_1^3 \,du = \frac{26}{6} \int_1^2 \frac{1}{u^2} \, du = \frac{26}{6} \left [ -\frac{1}{u} \right ]_1^2 = \frac{26}{6} \frac{1}{2} = \frac{26}{12} = \frac{13}{6}$$

### 3.2Accuracy and discussion¶

Wolfram Alpha confirms the final result, given that the integral/limits in terms of $u$ and $v$ are correct.

## 4Question 4¶

Find the $z$-coordinate of the centroid of the part $S$ of the spherical surface $x^2+y^2+z^2=4$ above $z=1$, i.e. compute

$$\bar z = \frac{\displaystyle \iint_S z\,dS}{\displaystyle \iint_S \,dS}$$

### 4.1Solution¶

The parametrisation for the sphere $x^2+y^2+z^2=2^2$ is as follows:

$$r(\theta, \phi) = 2\sin\phi \cos \theta \hat i + 2 \sin \phi \sin \theta \hat j + 2\cos \phi \hat k$$

where $\theta$ goes from 0 to $2\pi$ and $\phi$ goes from 0 to a specific angle $\phi_0$. To find this angle $\phi_0$, we draw a line from the origin to the point in the first quadrant (above the xy-plane and to the right of the y-axis) where the sphere intersects the plane $z=1$. The distance from the origin to the plane $z=1$ is 1, and the distance between the y-axis and this point is $\sqrt{3}$ (get this by substituting $z=1$ in the formula for the sphere) and so the length of the line segment is 2. Consequently, $\cos\phi_0 = \frac{1}{2}$. So $\cos\phi = \frac{\pi}{3}$. (If you don't know that offhand, you can probably derive the result by drawing out the unit circle or something.)

Anyways, now that you have the bounds, you still have to figure out the integrand. Basically you evaluate the following cross-product $r_{\theta} \times r_{\phi}$, which should give you $4\sin\phi$ (skipping all the intermediate steps here). This is the element of the surface integral, and also the Jacobian when converting into sphericals, oddly enough (why?). We also know that $z = 2\cos\phi$, from the parametrisation for the sphere. So the numerator becomes:

$$\iint_S z\,dS = \int_0^{2\pi} \int_0^{\pi/3} 2\cos\theta \cdot 4\sin \phi \,d\phi \,d\theta = 16\pi \int_0^{\pi/3} \cos \phi \sin \phi \,d\phi = 8\pi \int_0^{\pi/3} \sin(2\phi) \,d\phi = 4\pi \left [ -\cos(2\phi) \right ]_0^{\pi/3} = -4\pi \cdot (-3/2) =6\pi$$

(We can use the double angle formula for $\sin$, $2\sin\phi\cos\phi = \sin(2\phi)$, to simplify the integrand.)

The denominator is a bit simpler:

$$\iint_S \,dS = \int_0^{2\pi} \int_0^{\pi/3} 4\sin\phi \,d\phi \,d\theta = -8\pi \left [ \cos\phi \right ]_0^{\pi/3} \,d\phi = -8\pi \cdot (-1/2) = 4\pi$$

So the z-coordinate of the centroid is given by $$\displaystyle \frac{6\pi}{4\pi} = \frac{3}{2}$$

### 4.2Accuracy and discussion¶

Wolfram confirms the answer, given that the integrands and limits are correct. Are the integrands and limits correct? I think so.

## 5Question 5¶

Use Green's theorem to evaluate $\displaystyle \int_C (y-\sin x \sin y)\,dx - (x-\cos x \cos y)\,dy$ where $C$ is the boundary of the triangle with vertices (0,0), (1, 0), (0, 1) taken counterclockwise.

### 5.1Solution¶

First, we should revisit Green's theorem:

$$\int_C P\,dx + Q \,dy = \iint \left ( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right )\,dA$$

or, equivalently,

$$\int_C F \cdot ds = \iint \left ( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right )\,dA$$

(The two are equivalent because $F = P \hat i + Q \hat j$ and $ds = dx \hat i + dy\hat j$. It's probably obvious to you but it wasn't obvious to me so I'm writing it down in case I forget again.)

In any case, the values of $P$ and $Q$ that we're given in this problem are:

$$P = y-\sin x \sin y \quad Q = -x + \cos x \cos y$$

Now we just find the partial derivatives:

$$\frac{\partial Q}{\partial x} = -1 - \cos y \sin x \quad \frac{\partial P}{\partial y} = 1 - \sin x \cos y$$

So the double integral we have to solve is:

$$\iint_C -1 - \cos y\sin x - 1 + \sin x \cos y \,dA = \iint_C -2 \,dA$$

which is just -2 times the area of the triangle, which is $\displaystyle \frac{1}{2}$, so the final answer is -1.

### 5.2Accuracy and discussion¶

Seems to make sense. No guarantees though.

## 6Question 6¶

Evaluate $\displaystyle \int_C \vec F \cdot \,d\vec r$ where $\displaystyle \vec F = \frac{1}{x^2+y^2}(-y\hat i + x \hat j)$ and $C$ is the boundary of the square $-1\leq x \leq 1$, $-1\leq y \leq 1$ taken counterclockwise.

### 6.1Solution¶

This is a straightforward line integral. First, let's rewrite $F$ just for fun:

$$\vec F = -\frac{y}{x^2+y^2}\hat i + \frac{x}{x^2+y^2} \hat j$$

Now, let's parametrise $C$. We can split it up into four sections, obviously, since it's a square. Starting with the section on the top $C_1$ and moving around counterclockwise:

\begin{align*}C_1: \,x = -t, y = 1 & \quad -1 \leq t \leq 1 \quad \therefore r(t) = -t\hat i + \hat j, \, r'(t) = -\hat i\\ C_2: \, x = -1, y = -t & \quad -1 \leq t \leq 1 \quad \therefore r(t) = -\hat i -t\hat j, \, r'(t) = -\hat j\\ C_3: \,x = t, y = -1 & \quad -1 \leq t \leq 1 \quad \therefore r(t) = t\hat i - \hat j,\,r'(t) = \hat i\\ C_4: \, x = 1, y = t & \quad -1 \leq t \leq 1 \quad \therefore r(t) = \hat i + t\hat j,\,r'(t) = \hat j\end{align*}

Now we can evaluate the line integral along each segment, starting with $C_1$:

\begin{align*}\int_{C_1} \vec F \cdot \,d\vec r & = \int_{-1}^1 F(r(t)) \cdot r'(t) \,dt = \int_{-1}^1 \left (-\frac{1}{t^2+1} \hat i + \frac{-t}{t^2+1}\hat j \right ) \cdot (-\hat i) \,dt = \int_{-1}^1 \frac{1}{t^2+1} \\ \int_{C_2} \vec F \cdot \,d\vec r & = \int_{-1}^1 \left (\frac{t}{t^2+1} \hat i + \frac{1}{t^2+1}\hat j \right ) \cdot (-\hat j) \,dt = \int_{-1}^1 \frac{1}{t^2+1} \\ \int_{C_3} \vec F \cdot \,d\vec r & = \int_{-1}^1 \left (\frac{1}{t^2+1} \hat i + \frac{t}{t^2+1}\hat j \right ) \cdot (\hat i) \,dt = \int_{-1}^1 \frac{1}{t^2+1} \\ \int_{C_4} \vec F \cdot \,d\vec r & = \int_{-1}^1 \left (\frac{-t}{t^2+1} \hat i + \frac{1}{t^2+1}\hat j \right ) \cdot (\hat j) \,dt = \int_{-1}^1 \frac{1}{t^2+1} \end{align*}

(Sadly nothing cancelled out.) The integrand is therefore $\displaystyle\frac{4}{t^2+1}$. I suppose we could integrate that using trigonometric substitution but I'm not going to do that because I don't want to and Wolfram tells me the answer is $2\pi$ and really who wouldn't believe Wolfram Alpha?

### 6.2Accuracy and discussion¶

Not completely sure that the method is correct. It does feel right though. Also the integral at the end is less trivial than I would have thought it would be, but maybe my definition of a trivial integral is just off.

## 7Question 7¶

Let $\vec F = 7x\hat i - z\hat k$. Compute the flux of $\vec F$

(a) through the spherical surface $x^2+y^2+z^2=4$ with normal directed outwards, using the divergence theorem;

(b) through the hemispherical surface $x^2+y^2+z^2=4$, $z \geq 0$ (no bottom!) with normal directed upwards.

### 7.1Solution¶

(a)

Here's the formula for flux:

$$\iint_S F \cdot \hat n\,dS \text{ or } \iint_S F \cdot dS$$

Here's the divergence theorem:

$$\iint_S F \cdot dS = \iiint_E \text{div} F \,dV$$

where $E$ is the 3-dimensional region of which $S$ is the boundary surface. From the looks of things, we can find the flux simply by taking the triple integral of the divergence over the region $E$. So let's first find the divergence of the field $F = 7x\hat i - z\hat k$:

$$\text{div} F = 7 - 1 = 6$$

We could then take the triple integral, or we could just find the volume of the sphere and multiply that by 6 (note that this only works when the integrand is a constant - otherwise, you would have to go through the whole integration, most likely in spherical coordinates). Since the radius is 2, the volume is given by
$$V = \frac{4\pi r^3}{3} = \frac{32\pi}{3}$$

So the answer is a nice and clean $64\pi$.

(b)

Theoretically this would be just half the result from part (a), so $32\pi$. Not completely sure though, there may be something more subtle that I'm missing.

### 7.2Accuracy and discussion¶

The first one seems too simple to be correct, but question 6 (b) of the Winter 2010 final is similar and solutions are available for that one, so it's probably correct. Not sure about (b).

## 8Question 8¶

Use Stokes' theorem to calculate the line integral $\displaystyle \int_C -y^2\,dx + x^2\,dy - z^2\,dz$, where $C$ is the intersection of the surfaces $x^2+y^2=4y$ and $x+y+z=1$, with the orientation of $C$ counterclockwise when projected on the $xy$-plane.

### 8.1Solution¶

Stokes' theorem looks like this:

$$\int_C F \cdot d\vec r = \iint_S \text{curl } F \,d\vec S = \iint_S \text{curl } F \cdot \hat n \,dS$$

So let's first find the curl of $F$:

$$\text{curl } F = \det \begin{pmatrix} \hat i & \hat j & \hat k \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ -y^2 & x^2 & -z^2 \end{pmatrix} = (2x + 2y)\hat k$$

Now, using Stokes' theorem, we can transform the original problem into the following:

Evaluate the flux integral $$\iint_S G \cdot d\vec S$$, where $S$ is the part of the plane $x+y+z=1$ inside the cylinder $x^2+y^2=4y$, oriented with normals pointing upward.

lol