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Calculus 3 final exam review

20 practice problems across the topics a Calculus 3 final usually covers, in course order, each with a worked solution checked line by line. Your own exam may differ, so check your syllabus.

Try each problem on paper before opening its solution. If one gives you trouble, the topic link above it has more like it. How we verify

1. Dot product and angles

Problem 9.1

easy
Let \( \displaystyle \mathbf{u} = \langle 1, 3, 0 \rangle \) and \( \displaystyle \mathbf{v} = \langle 3, -4, -5 \rangle \). Find \( \displaystyle \mathbf{u} \cdot \mathbf{v} \) and the angle between them.
  1. \[ 3 \left(-4\right) + 0 \left(-5\right) + 1 \cdot 3 = -9 \]
    Multiply componentwise and add.✓ Proved
  2. \[ \sqrt{10} \]
    |u|.Reviewed
  3. \[ 5 \sqrt{2} \]
    |v|.Reviewed
  4. \[ - \frac{9 \sqrt{5}}{50} \]
    cos θ = u·v / (|u||v|).Reviewed
  5. θ = arccos of that.
    Reviewed
Answer \( \mathbf{u}\cdot\mathbf{v} = -9,\quad \theta = \operatorname{acos}{\left(- \frac{9 \sqrt{5}}{50} \right)} \approx 113.73^\circ \)

✓ Nihil obstat Lines: 1 proved, 4 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

2. Cross product

Problem 9.23

medium
Find \( \displaystyle \mathbf{u} \times \mathbf{v} \) for \( \displaystyle \mathbf{u} = \langle -1, 2, -2 \rangle,\ \mathbf{v} = \langle 2, -4, -4 \rangle \), and the area of the parallelogram they span.
  1. Expand the determinant with rows i j k, u, v.
    Reviewed
  2. \[ \left[\begin{matrix}-16\\-8\\0\end{matrix}\right] \]
    The components.Reviewed
  3. \[ 8 \sqrt{5} \]
    Area = |u × v|.Reviewed
Answer \( \mathbf{u}\times\mathbf{v} = \langle -16, -8, 0 \rangle,\quad \text{area} = 8 \sqrt{5} \)

✓ Nihil obstat Lines: 3 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

3. Equation of a plane

Problem 9.31

hard
Find an equation of the plane through the points \( \displaystyle (1, 2, 3),\ (-1, -1, 3),\ (-1, 3, 0) \).
  1. \[ \left[\begin{matrix}-2\\-3\\0\end{matrix}\right] \]
    PQ.Reviewed
  2. \[ \left[\begin{matrix}-2\\1\\-3\end{matrix}\right] \]
    PR.Reviewed
  3. \[ \left[\begin{matrix}9\\-6\\-8\end{matrix}\right] \]
    A normal vector n = PQ × PR (scaled).Reviewed
  4. \[ -27 \]
    n · P gives the constant.Reviewed
  5. The plane is 9*x - 6*y - 8*z = -27.
    Reviewed
Answer \( 9 x - 6 y - 8 z = -27 \)

✓ Nihil obstat Lines: 5 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

4. Lines in space

Problem 9.221

easy
Find parametric and symmetric equations of the line through \( \displaystyle P(6, 5, 2) \) and \( \displaystyle Q(-4, -4, -1) \).
  1. \[ \left[\begin{matrix}-10\\-9\\-3\end{matrix}\right] \]
    A direction vector.Reviewed
  2. r(t) = P + t·(Q − P); solving each coordinate for t gives the symmetric form.
    Reviewed
  3. \[ \left[\begin{matrix}-4\\-4\\-1\end{matrix}\right] \]
    At t = 1 the line reaches Q.Reviewed
Answer \( x = 6 - 10 t,\ y = 5 - 9 t,\ z = 2 - 3 t;\quad \frac{x - (6)}{-10} = \frac{y - (5)}{-9} = \frac{z - (2)}{-3} \)

✓ Nihil obstat Lines: 3 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

5. Velocity, acceleration and speed in space

Problem 9.232

medium
A particle has acceleration \( \displaystyle \mathbf a(t) = \left\langle \cos{\left(t \right)}, - \sin{\left(t \right)}, 1 \right\rangle \), initial velocity \( \displaystyle \left\langle -1, 0, -2 \right\rangle \) and initial position \( \displaystyle \left\langle 1, 2, -2 \right\rangle \). Find \( \displaystyle \mathbf r(t) \).
  1. \[ \left[\begin{matrix}\frac{d}{d t} \left(\sin{\left(t \right)} - 1\right)\\\frac{d}{d t} \left(\cos{\left(t \right)} - 1\right)\\\frac{d}{d t} \left(t - 2\right)\end{matrix}\right] = \left[\begin{matrix}\cos{\left(t \right)}\\- \sin{\left(t \right)}\\1\end{matrix}\right] \]
    Integrating a and using v(0) gives v(t) = <sin(t) - 1, cos(t) - 1, t - 2>.✓ Proved
  2. \[ \left[\begin{matrix}\frac{d}{d t} \left(- t - \cos{\left(t \right)} + 2\right)\\\frac{d}{d t} \left(- t + \sin{\left(t \right)} + 2\right)\\\frac{d}{d t} \left(\frac{t^{2}}{2} - 2 t - 2\right)\end{matrix}\right] = \left[\begin{matrix}\sin{\left(t \right)} - 1\\\cos{\left(t \right)} - 1\\t - 2\end{matrix}\right] \]
    Integrate again and use r(0).✓ Proved
  3. \[ \left[\begin{matrix}1\\2\\-2\end{matrix}\right] \]
    r(0) is the given position.Reviewed
Answer \( \mathbf r(t) = \left\langle - t - \cos{\left(t \right)} + 2, - t + \sin{\left(t \right)} + 2, \frac{t^{2}}{2} - 2 t - 2 \right\rangle \)

✓ Nihil obstat Lines: 2 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

6. Arc length and curvature of space curves

Problem 9.656

medium
Find the curvature of \( \displaystyle \mathbf r(t) = \left\langle t, t^{2}, 0 \right\rangle \) at \( \displaystyle t = -1 \).
  1. \[ \left[\begin{matrix}\left. \frac{d}{d t} t \right|_{\substack{ t=-1 }}\\\left. \frac{d}{d t} t^{2} \right|_{\substack{ t=-1 }}\\\left. \frac{d}{d t} 0 \right|_{\substack{ t=-1 }}\end{matrix}\right] = \left[\begin{matrix}1\\-2\\0\end{matrix}\right] \]
    r′(-1).✓ Proved
  2. \[ \left[\begin{matrix}\left. \frac{d^{2}}{d t^{2}} t \right|_{\substack{ t=-1 }}\\\left. \frac{d^{2}}{d t^{2}} t^{2} \right|_{\substack{ t=-1 }}\\\left. \frac{d^{2}}{d t^{2}} 0 \right|_{\substack{ t=-1 }}\end{matrix}\right] = \left[\begin{matrix}0\\2\\0\end{matrix}\right] \]
    r″(-1).✓ Proved
  3. \[ \left[\begin{matrix}\left(-2\right) 0 + 2 \cdot 0\\0 \cdot 0 + 0\\0 \cdot 2 + 1 \cdot 2\end{matrix}\right] = \left[\begin{matrix}0\\0\\2\end{matrix}\right] \]
    r′ × r″.✓ Proved
  4. \[ \frac{\sqrt{0^{2} + 0^{2} + 2^{2}}}{\left(\sqrt{0^{2} + 1^{2} + \left(-2\right)^{2}}\right)^{3}} = \frac{2 \sqrt{5}}{25} \]
    κ = ‖r′ × r″‖ / ‖r′‖³.✓ Proved
Answer \( \kappa = \frac{2 \sqrt{5}}{25} \)

✓ Nihil obstat Every line of this solution was proved by the computer algebra system SymPy. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

7. Partial derivatives

Problem 10.1

hard
For \( \displaystyle f(x, y) = \ln{\left(x^{2} + y^{2} + 1 \right)} \), find \( \displaystyle f_x \), \( \displaystyle f_y \) and \( \displaystyle f_{xy} \).
  1. For f_x hold y constant; for f_y hold x constant.
    Reviewed
  2. \[ \frac{\partial}{\partial x} \ln{\left(x^{2} + y^{2} + 1 \right)} = \frac{2 x}{x^{2} + y^{2} + 1} \]
    f_x.✓ Proved
  3. \[ \frac{\partial}{\partial y} \ln{\left(x^{2} + y^{2} + 1 \right)} = \frac{2 y}{x^{2} + y^{2} + 1} \]
    f_y.✓ Proved
  4. \[ \frac{\partial}{\partial y} \frac{2 x}{x^{2} + y^{2} + 1} = - \frac{4 x y}{\left(x^{2} + y^{2} + 1\right)^{2}} \]
    f_xy: differentiate f_x with respect to y.✓ Proved
  5. \[ \frac{\partial}{\partial x} \frac{2 y}{x^{2} + y^{2} + 1} = - \frac{4 x y}{\left(x^{2} + y^{2} + 1\right)^{2}} \]
    Clairaut: f_yx is the same.✓ Proved
Answer \( f_x = \frac{2 x}{x^{2} + y^{2} + 1},\quad f_y = \frac{2 y}{x^{2} + y^{2} + 1},\quad f_{xy} = - \frac{4 x y}{\left(x^{2} + y^{2} + 1\right)^{2}} \)

✓ Nihil obstat Lines: 4 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

8. The multivariable chain rule

Problem 10.345

medium
Let \( \displaystyle z = x \cos{\left(y \right)} + y e^{x} \) with \( \displaystyle x = s t \), \( \displaystyle y = s + t \). Find \( \displaystyle \frac{\partial z}{\partial s} \) at \( \displaystyle s = 1 \), \( \displaystyle t = 1 \).
  1. \[ \left[\begin{matrix}\frac{\partial}{\partial x} \left(x \cos{\left(y \right)} + y e^{x}\right)\\\frac{\partial}{\partial y} \left(x \cos{\left(y \right)} + y e^{x}\right)\end{matrix}\right] = \left[\begin{matrix}y e^{x} + \cos{\left(y \right)}\\- x \sin{\left(y \right)} + e^{x}\end{matrix}\right] \]
    ∂z/∂x and ∂z/∂y.✓ Proved
  2. ∂z/∂s = (∂z/∂x)(∂x/∂s) + (∂z/∂y)(∂y/∂s).
    Reviewed
  3. \[ - \sin{\left(2 \right)} + \cos{\left(2 \right)} + 3 e = \sqrt{2} \cos{\left(\frac{\pi}{4} + 2 \right)} + 3 e \]
    Substitute.✓ Proved
Answer \( \sqrt{2} \cos{\left(\frac{\pi}{4} + 2 \right)} + 3 e \)

✓ Nihil obstat Lines: 2 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

9. Gradient and directional derivatives

Problem 10.9

medium Video
Find the directional derivative of \( \displaystyle f(x, y) = \sqrt{x^{2} + y^{2}} \) at \( \displaystyle (-1, 1) \) in the direction of \( \displaystyle \langle 1, 1 \rangle \).
  1. \[ \left[\begin{matrix}\frac{\partial}{\partial x} \sqrt{x^{2} + y^{2}}\\\frac{\partial}{\partial y} \sqrt{x^{2} + y^{2}}\end{matrix}\right] = \left[\begin{matrix}\frac{x}{\sqrt{x^{2} + y^{2}}}\\\frac{y}{\sqrt{x^{2} + y^{2}}}\end{matrix}\right] \]
    ∇f.✓ Proved
  2. \[ \left[\begin{matrix}\frac{\left(-1\right) \sqrt{2}}{2}\\\frac{\sqrt{2}}{2}\end{matrix}\right] = \left[\begin{matrix}- \frac{\sqrt{2}}{2}\\\frac{\sqrt{2}}{2}\end{matrix}\right] \]
    ∇f at the point.Reviewed
  3. \[ \left[\begin{matrix}\frac{\sqrt{2}}{2}\\\frac{\sqrt{2}}{2}\end{matrix}\right] \]
    The unit direction u.Reviewed
  4. \[ 0 \]
    D_u f = ∇f · u.Reviewed
Answer \( 0 \)
A contour map of f(x, y) = sqrt(x² + y²): each curve joins points where f has the same value. At (-1, 1) the gradient arrow points straight uphill, across the contours, and the direction of travel is drawn beside it.

✓ Nihil obstat Lines: 1 proved, 3 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

10. Tangent planes

Problem 10.87

medium
Find the tangent plane to \( \displaystyle z = e^{x - y} \) at the point \( \displaystyle (0, 0, 1) \).
  1. z = f(a,b) + f_x(a,b)(x − a) + f_y(a,b)(y − b).
    Reviewed
  2. \[ \left. \frac{\partial}{\partial x} e^{x - y} \right|_{\substack{ x=0\\ y=0 }} = 1 \]
    f_x at the point.✓ Proved
  3. \[ \left. \frac{\partial}{\partial y} e^{x - y} \right|_{\substack{ x=0\\ y=0 }} = -1 \]
    f_y at the point.✓ Proved
  4. \[ x - y + 1 \]
    The plane.Reviewed
Answer \( z = x - y + 1 \)

✓ Nihil obstat Lines: 2 proved, 2 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

11. Local extrema and saddle points

Problem 10.39

easy Video
Find and classify the critical points of \( \displaystyle f(x, y) = x^{2} - 2 x - y^{2} + 4 y - 3 \).
  1. \[ \left[\begin{matrix}\frac{\partial}{\partial x} \left(x^{2} - 2 x - y^{2} + 4 y - 3\right)\\\frac{\partial}{\partial y} \left(x^{2} - 2 x - y^{2} + 4 y - 3\right)\end{matrix}\right] = \left[\begin{matrix}2 x - 2\\4 - 2 y\end{matrix}\right] \]
    The partial derivatives.✓ Proved
  2. \[ \left[\begin{matrix}0\\0\end{matrix}\right] \]
    Both vanish at (1, 2), the only solution.Reviewed
  3. \[ -4 \]
    D = f_xx f_yy − f_xy².Reviewed
  4. D = -4 < 0: a saddle point.
    Reviewed
Answer \( \text{saddle point at } (1, 2) \)
A contour map of f(x, y) = x² - 2x - y² + 4y - 3: each curve joins points where f has the same value. The contours close in around the critical point (1, 2).

✓ Nihil obstat Lines: 1 proved, 3 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

12. Lagrange multipliers

Problem 10.55

medium Video
Use Lagrange multipliers to find the maximum and minimum of \( \displaystyle f(x, y) = 4 x + y \) on the circle \( \displaystyle x^2 + y^2 = 25 \).
  1. Solve ∇f = λ∇g with g(x, y) = x² + y² − r² = 0.
    Reviewed
  2. \[ \left[\begin{matrix}\frac{\partial}{\partial x} \left(4 x + y\right)\\\frac{\partial}{\partial y} \left(4 x + y\right)\end{matrix}\right] = \left[\begin{matrix}4\\1\end{matrix}\right] \]
    ∇f.✓ Proved
  3. \[ \left[\begin{matrix}\frac{\partial}{\partial x} \left(x^{2} + y^{2} - 25\right)\\\frac{\partial}{\partial y} \left(x^{2} + y^{2} - 25\right)\end{matrix}\right] = \left[\begin{matrix}2 x\\2 y\end{matrix}\right] \]
    ∇g.✓ Proved
  4. So x = 4/(2λ), y = 1/(2λ); substituting into the constraint gives λ = ±√(17)/(2√25).
    Reviewed
  5. \[ 5 \sqrt{17} \]
    The maximum; the minimum is its negative.Reviewed
Answer \( \max = 5 \sqrt{17},\ \min = - 5 \sqrt{17} \)
A contour map of f(x, y) = 4x + y: each curve joins points where f has the same value. The constraint circle of radius 5 is drawn over it; the largest and smallest values on the circle are where a level line just touches it.

✓ Nihil obstat Lines: 2 proved, 3 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

13. Double integrals over general regions

Problem 11.26

medium
Evaluate \( \displaystyle \iint_D x^{2} \, dA \) where D is bounded by \( \displaystyle y = 0 \), \( \displaystyle y = \sqrt{x} \), \( \displaystyle x = 0 \) and \( \displaystyle x = 2 \).
  1. D is Type I: for each x in [0, 2], y runs from 0 to sqrt(x).
    Reviewed
  2. \[ \int\limits_{0}^{\sqrt{x}} x^{2}\, dy = x^{\frac{5}{2}} \]
    The inner integral.✓ Proved
  3. \[ \int\limits_{0}^{2} x^{\frac{5}{2}}\, dx = \frac{16 \sqrt{2}}{7} \]
    The outer integral.✓ Proved
Answer \( \frac{16 \sqrt{2}}{7} \)

✓ Nihil obstat Lines: 2 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

14. Double integrals in polar coordinates

Problem 11.33

medium
Evaluate \( \displaystyle \iint_D e^{- x^{2} - y^{2}} \, dA \) over the disk \( \displaystyle x^2 + y^2 \le 4 \) using polar coordinates.
  1. In polar coordinates the disk is 0 ≤ r ≤ R, 0 ≤ θ ≤ 2π, and dA = r dr dθ.
    Reviewed
  2. \[ e^{- r^{2} \sin^{2}{\left(\theta \right)} - r^{2} \cos^{2}{\left(\theta \right)}} = e^{- r^{2}} \]
    The integrand in r and θ.✓ Proved
  3. \[ \int\limits_{0}^{2 \pi}\int\limits_{0}^{2} r e^{- r^{2}}\, dr\, d\theta = - \frac{\pi}{e^{4}} + \pi \]
    Integrate, remembering the extra r.✓ Proved
Answer \( - \frac{\pi}{e^{4}} + \pi \)

✓ Nihil obstat Lines: 2 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

15. Triple integrals

Problem 11.43

easy
Evaluate \( \displaystyle \iiint_B x^{2} z \, dV \) where \( \displaystyle B = [0, 2] \times [0, 1] \times [0, 2] \).
  1. \[ \int\limits_{0}^{2} x^{2} z\, dz = 2 x^{2} \]
    Integrate in z.✓ Proved
  2. \[ \int\limits_{0}^{1} 2 x^{2}\, dy = 2 x^{2} \]
    Then y.✓ Proved
  3. \[ \int\limits_{0}^{2} 2 x^{2}\, dx = \frac{16}{3} \]
    Then x.✓ Proved
Answer \( \frac{16}{3} \)

✓ Nihil obstat Every line of this solution was proved by the computer algebra system SymPy. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

16. Triple integrals in spherical coordinates

Problem 11.225

easy
Use spherical coordinates to find the volume of the upper half of the ball \( \displaystyle x^2 + y^2 + z^2 \le 4 \).
  1. ρ runs from 0 to 2, φ from 0 to pi/2, θ around the full circle; dV = ρ² sin φ dρ dφ dθ.
    Reviewed
  2. \[ \int\limits_{0}^{2 \pi}\int\limits_{0}^{\frac{\pi}{2}}\int\limits_{0}^{2} \rho^{2} \sin{\left(\phi \right)}\, d\rho\, d\phi\, d\theta = \frac{16 \pi}{3} \]
    The iterated integral factors into three one-variable integrals.✓ Proved
Answer \( \frac{16 \pi}{3} \approx 16.75516 \)

✓ Nihil obstat Lines: 1 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

17. Line integrals and work

Problem 12.2

medium
Find the work done by \( \displaystyle \mathbf{F} = \langle - y, x \rangle \) along the unit circle, counterclockwise.
  1. Parametrize: r(t) = ⟨cos(t), sin(t)⟩ for 0 ≤ t ≤ 2*pi.
    Reviewed
  2. \[ \left[\begin{matrix}\frac{d}{d t} \cos{\left(t \right)}\\\frac{d}{d t} \sin{\left(t \right)}\end{matrix}\right] = \left[\begin{matrix}- \sin{\left(t \right)}\\\cos{\left(t \right)}\end{matrix}\right] \]
    r'(t).✓ Proved
  3. \[ \sin^{2}{\left(t \right)} + \cos^{2}{\left(t \right)} = 1 \]
    F(r(t)) · r'(t).✓ Proved
  4. \[ \int\limits_{0}^{2 \pi} 1\, dt = 2 \pi \]
    W = ∫ F · dr.✓ Proved
Answer \( 2 \pi \)

✓ Nihil obstat Lines: 3 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

18. Green's theorem

Problem 12.16

easy
Use Green's theorem to evaluate \( \displaystyle \oint_C (x^{2} - y)\,dx + (x + y^{2})\,dy \), where C is the boundary of the rectangle \( \displaystyle [0, 3] \times [0, 1] \), counterclockwise.
  1. Green's theorem: ∮ P dx + Q dy = ∬ (∂Q/∂x − ∂P/∂y) dA.
    Reviewed
  2. \[ \frac{\partial}{\partial x} \left(x + y^{2}\right) - \frac{\partial}{\partial y} \left(x^{2} - y\right) = 2 \]
    ∂Q/∂x − ∂P/∂y.✓ Proved
  3. \[ \int\limits_{0}^{1}\int\limits_{0}^{3} 2\, dx\, dy = 6 \]
    Integrate over the rectangle.✓ Proved
Answer \( 6 \)

✓ Nihil obstat Lines: 2 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

19. The divergence theorem

Problem 12.208

medium
Use the divergence theorem to find the outward flux of \( \displaystyle \mathbf F = \left(y z\right)\mathbf i + \left(- 2 y\right)\mathbf j + \left(0\right)\mathbf k \) across the sphere \( \displaystyle x^2 + y^2 + z^2 = 1 \).
  1. \[ \frac{d}{d z} 0 + \frac{d}{d y} \left(- 2 y\right) + \frac{\partial}{\partial x} y z = -2 \]
    div F.✓ Proved
  2. Flux out of a closed surface = ∭ div F dV over the solid inside it (in spherical coordinates).
    Reviewed
  3. \[ \int\limits_{0}^{2 \pi}\int\limits_{0}^{\pi}\int\limits_{0}^{1} \left(- 2 \rho^{2} \sin{\left(\phi \right)}\right)\, d\rho\, d\phi\, d\theta = - \frac{8 \pi}{3} \]
    The triple integral.✓ Proved
Answer \( - \frac{8 \pi}{3} \)

✓ Nihil obstat Lines: 2 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

20. Stokes' theorem

Problem 12.370

hard
Use Stokes' theorem to evaluate \( \displaystyle \oint_C \mathbf F \cdot d\mathbf r \) for \( \displaystyle \mathbf F = \left\langle - y, x, z \right\rangle \), where \( \displaystyle C \) is the circle \( \displaystyle x^2 + y^2 = 4 \) in the plane \( \displaystyle z = 0 \), counterclockwise seen from above.
  1. Take S to be the flat disk the circle bounds, with upward normal k. Stokes: ∮_C F · dr = ∬_S (curl F) · k dA.
    Reviewed
  2. \[ \left[\begin{matrix}- \frac{d}{d z} x + \frac{d}{d y} z\\\frac{d}{d z} \left(- y\right) - \frac{d}{d x} z\\\frac{d}{d x} x - \frac{d}{d y} \left(- y\right)\end{matrix}\right] = \left[\begin{matrix}0\\0\\2\end{matrix}\right] \]
    curl F.✓ Proved
  3. \[ \int\limits_{0}^{2 \pi}\int\limits_{0}^{2} r \left. 2 \right|_{\substack{ x=r \cos{\left(\theta \right)}\\ y=r \sin{\left(\theta \right)} }}\, dr\, d\theta = 8 \pi \]
    The k-component of the curl over the disk, in polar coordinates.✓ Proved
  4. \[ \int\limits_{0}^{2 \pi} 4\, dt = 8 \pi \]
    The circulation computed directly around C agrees.✓ Proved
Answer \( 8 \pi \)

✓ Nihil obstat Lines: 3 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

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