Free Calculus III: Multivariable Calculus Study Resources

To correctly set up a double integral in polar coordinates over a general region $D$, we must analyze three structural components:

1. **The Differential Area Element ($dA):** In polar coordinates, the transformation from Cartesian coordinates results in dA = r \, dr \, d\theta. Option B is incorrect because it uses the Cartesian-like differential dr \, d\theta and omits the Jacobian factor r$.

2. The Order of Integration and Limits: For a region defined by $h_1(\theta) \le r \le h_2(\theta)$ and $\alpha \le \theta \le \beta$, we must perform the inner integration with respect to $r because its bounds are functions of the outer variable \theta. The outer integration must have constant limits (\alpha$ and $\beta$) to ensure the final result is a scalar. Option C is incorrect because it places functional limits on the outer integral, which is mathematically invalid for an iterated integral.

3. Placement of the Jacobian $r$: The factor $r is part of the integrand for both the inner and outer integrals. In Option D, r is moved outside the inner integral with respect to dr$. Since $r$ is the variable of integration for the inner integral, it cannot be treated as a constant and factored out of that step.

Therefore, the only structurally valid and correct setup is Option A: $\int_{\alpha}^{\beta} \int_{h_1(\theta)}^{h_2(\theta)} f(r\cos\theta, r\sin\theta) \cdot r \, dr \, d\theta$

To transform an integral into spherical coordinates, we must account for the change in volume scales using the Jacobian determinant. The transformation equations are:

$x = \rho \sin \phi \cos \theta$ $y = \rho \sin \phi \sin \theta$ $z = \rho \cos \phi$

The absolute value of the Jacobian for this transformation is $|J| = \rho^2 \sin \phi$. Consequently, the differential volume element is:

$dV = \rho^2 \sin \phi \, d\rho \, d\phi \, d\theta$

• Option B is the differential volume element for cylindrical coordinates.
• Option C incorrectly omits the square on the radial distance ($\rho^2$).
• Option D incorrectly uses the cosine function; the projection onto the $xy-plane requires the sine of the polar angle \phi$.

The correct choice is A.

To evaluate the surface integral $\iint_S (\nabla \times \mathbf{F}) \cdot d\mathbf{S}$ for this complex piecewise surface, we apply Stokes' Theorem, which relates the surface integral of the curl to the line integral over the boundary curve $C = \partial S$:

$\iint_S (\nabla \times \mathbf{F}) \cdot d\mathbf{S} = \oint_C \mathbf{F} \cdot d\mathbf{r}$

1. Identify the Boundary Curve $C$ The surface $S is open at the base z=0. The boundary curve C is the circle defined by x^2 + y^2 = 9$ in the plane $z=0$.

2. Determine the Orientation of $C$ The surface is oriented with outward-pointing normal vectors. By the right-hand rule, walking along the boundary at $z=0 with the surface (the cylinder walls) on the left requires a **counter-clockwise** traversal when viewed from above. We parameterize C$ as: $\mathbf{r}(t) = \langle 3\cos t, 3\sin t, 0 \rangle, \quad t \in [0, 2\pi]$ Thus, $d\mathbf{r} = \langle -3\sin t, 3\cos t, 0 \rangle dt$.

3. Evaluate the Line Integral On the boundary curve $C$, $z = 0. Substituting this into the vector field \mathbf{F}$: $\mathbf{F}(x, y, 0) = \langle y + e^{0^2}, x^2 + \ln(1 + 0^2), 0 \rangle = \langle y + 1, x^2, 0 \rangle$ Now substitute the parameterization $x = 3\cos t$ and $y = 3\sin t$: $\mathbf{F}(\mathbf{r}(t)) = \langle 3\sin t + 1, 9\cos^2 t, 0 \rangle$ Compute the dot product $\mathbf{F} \cdot d\mathbf{r}$: $\mathbf{F} \cdot d\mathbf{r} = (3\sin t + 1)(-3\sin t) + (9\cos^2 t)(3\cos t) + (0)(0)$ $\mathbf{F} \cdot d\mathbf{r} = -9\sin^2 t - 3\sin t + 27\cos^3 t$ Integrating from 0 to $2\pi$:

• $\int_0^{2\pi} -9\sin^2 t \, dt = -9\pi$
• $\int_0^{2\pi} -3\sin t \, dt = 0$
• $\int_0^{2\pi} 27\cos^3 t \, dt = 0$

The total integral is $-9\pi$.

Clairaut's Theorem (also known as Schwarz's Theorem) states that if the partial derivatives of a function are continuous on an open set, then the order of differentiation does not matter; only the total number of times the function is differentiated with respect to each variable determines the value of the mixed partial derivative.

1. **Evaluate $f_{xyx}**: This denotes the third-order partial derivative where the function is differentiated with respect to x$ twice and $y$ once (specifically, $\frac{\partial}{\partial x}(\frac{\partial}{\partial y}(\frac{\partial f}{\partial x}))$, depending on the convention).
2. **Evaluate $f_{xxy}**: This also denotes a third-order partial derivative where the function is differentiated with respect to x$ twice and $y$ once (specifically, $\frac{\partial}{\partial y}(\frac{\partial}{\partial x}(\frac{\partial f}{\partial x}))$).

Since the set of variables differentiated ($x, x, y) is identical in both cases and the derivatives are continuous, the extension of Clairaut's Theorem guarantees that **f_{xyx} = f_{xxy}$**.

• Option B is incorrect because the counts of variables differ ($x$ twice, $y$ once vs. $x$ once, $y$ twice).
• Option C is incorrect because it compares differentiation with respect to different variables ($x, y, z$ vs. $x, z, z$).
• Option D is incorrect as there is no sign change (negation) involved in the equality of mixed partials for continuous functions.

To determine the torque $\vec{\tau}$, $we must calculate the cross product of the position vector \vec{r}$ and $the force vector \vec{F} in the specific order: \vec{\tau} = \vec{r} \times \vec{F}$.

1. Set up the determinant: $\vec{\tau} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & -1 & 3 \\ 4 & 5 & -2 \end{vmatrix}$

2. Expand the determinant:

• $\mathbf{i}$-component: (-1)(-2) - (3)(5) = 2 - 15 = -13
• $\mathbf{j}$-component: -[(2)(-2) - (3)(4)] = -[-4 - 12] = -[-16] = 16
• $\mathbf{k}$-component: (2)(5) - (-1)(4) = 10 + 4 = 14

This yields the vector $\langle -13, 16, 14 \rangle$.

1. Analysis of distractors:

• Option B represents $\vec{F} \times \vec{r}$. Because the cross product is anti-commutative, reversing the order flips the signs of all components.
• Option C occurs if one fails to apply the negative sign to the $\mathbf{j}$-component cofactor expansion.
• Option D is the result of component-wise multiplication (Hadamard product), which has no physical meaning in this context.

1. Verification: A valid torque vector must be orthogonal to both $\vec{r}$ and $\vec{F}$. $\vec{\tau} \cdot \vec{r} = (-13)(2) + (16)(-1) + (14)(3) = -26 - 16 + 42 = 0$.

Therefore, the correct torque vector is $\vec{\tau} = \langle -13, 16, 14 \rangle$ Nm.

To find the velocity and acceleration vectors, we must compute the first and second derivatives of the position vector $\mathbf{r}(t) = \langle \cos(t^2), \sin(t^2), t^2 \rangle$.

1. Calculate the velocity vector $\mathbf{v}(t)$: Using the chain rule, $\mathbf{v}(t) = \mathbf{r}'(t) = \langle -2t \sin(t^2), 2t \cos(t^2), 2t \rangle$. At $t = \sqrt{\pi}$, we have $t^2 = \pi$. Thus: $\mathbf{v}(\sqrt{\pi}) = \langle -2\sqrt{\pi} \sin(\pi), 2\sqrt{\pi} \cos(\pi), 2\sqrt{\pi} \rangle = \langle 0, -2\sqrt{\pi}, 2\sqrt{\pi} \rangle$.

2. Calculate the acceleration vector $\mathbf{a}(t)$: Using the product and chain rules, $\mathbf{a}(t) = \mathbf{v}'(t) = \langle -2\sin(t^2) - 4t^2\cos(t^2), 2\cos(t^2) - 4t^2\sin(t^2), 2 \rangle$. At $t = \sqrt{\pi}$: $x''(t) = -2\sin(\pi) - 4\pi\cos(\pi) = 0 - 4\pi(-1) = 4\pi$ $y''(t) = 2\cos(\pi) - 4\pi\sin(\pi) = 2(-1) - 0 = -2$ $z''(t) = 2$ So, $\mathbf{a}(\sqrt{\pi}) = \langle 4\pi, -2, 2 \rangle$.

Comparing these to the options, the correct choice is A.

To identify the correct trial form for the particular solution $y_p(x)$ using the method of undetermined coefficients, follow these steps:

1. Identify the nonhomogeneous term: The term on the right-hand side is $g(x) = 4x^2 - 1$.
2. Determine the function type: $g(x)$ is a polynomial of degree 2.
3. Apply the general rule: If $g(x) is a polynomial of degree n, the trial solution must be a general polynomial of the same degree n. This means we must include all powers of x$ from $n down to 0, even if those terms are not present in g(x)$.
4. **Formulate $y_p(x)**: For a degree 2 polynomial, the general form is Ax^2 + Bx + C$.
5. Check for overlap: The characteristic equation of the homogeneous side is $r^2 + 3r + 2 = 0$, giving roots $r = -1$ and $r = -2. The homogeneous solutions are c_1e^{-x}$ and $c_2e^{-2x}$. Since there is no overlap between the polynomial trial form and the homogeneous solutions, no modification is needed.

Therefore, the correct trial form is $y_p(x) = Ax^2 + Bx + C$.

To define the binormal vector $B(t)$ in the Frenet-Serret frame:

1. Understand the Orthogonality: The vectors $T(t)$, $N(t)$, and $B(t) form an orthonormal basis for \mathbb{R}^3$ at any point along a smooth curve.
2. Apply the Cross Product: By convention, the binormal vector is defined as the cross product of the unit tangent and the principal unit normal vectors: $B(t) = T(t) \times N(t)$
3. Right-Hand Rule: This specific order ($T$ then $N$) ensures that the frame follows the right-hand rule.
4. Evaluate Distractors:
   • Option B ($N \times T$) yields $-B(t)$, which is the opposite direction.
   • Option C is a dot product, which results in a scalar (0, since they are orthogonal), not a vector.
   • Option D refers to the derivative of the normal vector, which relates to torsion and the binormal vector via the Frenet-Serret formulas but is not the definition of $B(t)$ itself.

Therefore, the correct definition is $B(t) = T(t) \times N(t)$.

In a three-dimensional coordinate system, the coordinate planes are the surfaces where one of the coordinates is always zero. The name of the plane tells you which variables are allowed to vary; the variable not mentioned in the name is the one that must equal zero.

1. The $xz$-plane contains the $x$-axis and the $z$-axis.
2. For any point to lie within this plane, it cannot have any displacement along the $y$-axis.
3. Therefore, the equation for the $xz$-plane is $y = 0$.

By contrast:

• $x = 0$ defines the $yz$-plane.
• $z = 0$ defines the $xy$-plane.
• $x = 0$ and $z = 0$ describes the $y$-axis (a line), not a plane.

Curvature $\kappa measures the rate at which a curve changes direction at a point. By definition, the unit tangent vector \mathbf{T}$ has a constant magnitude of 1, so its derivative must be orthogonal to it and only represents a change in direction.

1. The fundamental definition of curvature is the magnitude of the rate of change of the unit tangent vector with respect to the distance traveled along the curve (arc length $s$): $\kappa = \left\| \frac{d\mathbf{T}}{ds} \right\|$
2. Option B is incorrect because $\left\| \frac{d\mathbf{T}}{dt} \right\|$ depends on the speed of the parametrization ($t$), whereas curvature is an intrinsic property of the curve itself.
3. Option C is the radius of curvature $\rho, which is the reciprocal of curvature (\rho = 1/\kappa$).
4. Option D is the magnitude of the acceleration vector, which includes both the change in speed (tangential) and the change in direction (normal).

The correct definition is $\kappa = \left\| \frac{d\mathbf{T}}{ds} \right\|$.

To analyze the curvature $\kappa, we use the general formula for a curve parameterized by t$: $\kappa(t) = \frac{||\mathbf{r}'(t) \times \mathbf{r}''(t)||}{||\mathbf{r}'(t)||^3}$

1. Compute the derivatives: $\mathbf{r}'(t) = \langle -2\sin(2t), 2\cos(2t), 4 \rangle$ $\mathbf{r}''(t) = \langle -4\cos(2t), -4\sin(2t), 0 \rangle$

2. Compute the magnitudes: $||\mathbf{r}'(t)|| = \sqrt{(-2\sin(2t))^2 + (2\cos(2t))^2 + 4^2} = \sqrt{4(\sin^2(2t) + \cos^2(2t)) + 16} = \sqrt{20} = 2\sqrt{5}$

3. Compute the cross product: $\mathbf{r}'(t) \times \mathbf{r}''(t) = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -2\sin(2t) & 2\cos(2t) & 4 \\ -4\cos(2t) & -4\sin(2t) & 0 \end{vmatrix}$ $= \langle 16\sin(2t), -16\cos(2t), 8\sin^2(2t) + 8\cos^2(2t) \rangle = \langle 16\sin(2t), -16\cos(2t), 8 \rangle$

4. Compute the magnitude of the cross product: $||\mathbf{r}'(t) \times \mathbf{r}''(t)|| = \sqrt{(16\sin(2t))^2 + (-16\cos(2t))^2 + 8^2} = \sqrt{256 + 64} = \sqrt{320} = 8\sqrt{5}$

5. Calculate curvature: $\kappa = \frac{8\sqrt{5}}{(2\sqrt{5})^3} = \frac{8\sqrt{5}}{8 \cdot 5\sqrt{5}} = \frac{1}{5}$

Analysis of properties: Curvature is an intrinsic property of the curve's geometry (the tightness of the bend) and does not depend on the speed of parameterization. While reparameterizing by arc length $s simplifies the formula to \kappa = ||d\mathbf{T}/ds||, the numerical value of \kappa at any point on the path remains \mathbf{1/5} regardless of the parameter used. Option D is incorrect because \kappa = ||\mathbf{r}''|| only holds if the curve is already parameterized by arc length (|\mathbf{r}'|=1$).

To convert the rectangular point $(x, y) = (-\sqrt{3}, -1)$ to polar coordinates $(r, \theta)$, we follow these steps:

1. Calculate the radial distance $r$: Using the formula $r^2 = x^2 + y^2$: $r^2 = (-\sqrt{3})^2 + (-1)^2 = 3 + 1 = 4$ Thus, $r$ can be 2 or $-2$.

2. Determine the angle $\theta$: Calculate the reference angle using $\tan \alpha = |y/x|$: $\tan \alpha = \frac{|-1|}{|-\sqrt{3}|} = \frac{1}{\sqrt{3}}$ $\alpha = \arctan\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6}$ Since the rectangular point $(-\sqrt{3}, -1) lies in **Quadrant III**, the standard angle \theta$ (where $r > 0$) is: $\theta = \pi + \frac{\pi}{6} = \frac{7\pi}{6}$ So, one representation is $(2, \frac{7\pi}{6})$.

3. Find equivalent representations with negative $r$: A point $(r, \theta)$ is equivalent to $(-r, \theta \pm \pi)$. Using $r = -2$, we adjust our standard angle: $\theta' = \frac{7\pi}{6} - \pi = \frac{\pi}{6}$ This gives us the pair $(-2, \frac{\pi}{6})$.

4. Evaluate the options:

• A is $(2, \frac{\pi}{6}), which is in Quadrant I (x=\sqrt{3}, y=1)$.
• B is $(-2, \frac{7\pi}{6})$. A negative radius at $7\pi/6$ (QIII) reflects the point back into Quadrant I.
• D is $(2, \frac{11\pi}{6})$, which is in Quadrant IV $(x=\sqrt{3}, y=-1)$.
• C is $(-2, \frac{\pi}{6})$. The angle $\pi/6 points toward Q1, but the negative radius reflects it 180^\circ$ into Q3 at $(-\sqrt{3}, -1)$.

To determine the arc length $s of a curve defined parametrically by x(t)$ and $y(t)$ on the interval $[a, b]$, we use the integral of the speed:

$s = \int_{a}^{b} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt$