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Course Notes for Multivariable Calculus

Remkes Kooistra

Section 10.3 Week 10 Activity

Subsection 10.3.1 Activity

Activity 10.3.1.

Calculate the line integral of \(F = (x^2 + y^2, 4xy) \) over the parametric curve \(\gamma = (t^2, 1-t) \) for \(t \in [0,2]\text{.}\)
Solution.
I need to evalute the field along the curve, by replacing the variables defining the field with the matching components of the curve.
\begin{align*} F(\gamma(t)) \amp = ((t^2)^2 + (1-t)^2, 4t^2(1-t)) = (t^4 + t^2 - 2t + 1, -4t^3 + 4t^2) \end{align*}
Next I need to calculate the tangent of the curve.
\begin{align*} \gamma^\prime(t) \amp = (2t, -1) \end{align*}
Then I need the dot product of the two previous calculations.
\begin{align*} F(\gamma(t)) \cdot \gamma^\prime(t) \amp = 2t(t^4+t^2-2t+1) - (-4t^3+4t^2) \\ \amp = 2t^4 + 2t^3 - 4t^2 + 2t + 4t^3 - 4t^2 \\ \amp = 2t^5 + 6t^3 - 8t^2 + 2t \end{align*}
Finally, I do the line integral by integrating this dot product over the parameter domain.
\begin{align*} \int_0^{2} F(\gamma(t)) \cdot \gamma^\prime(t) dt \amp = \int_0^2 2t^5 + 6t^3 - 8t^2 + 2t dt \\ \amp = \left( \frac{2t^6}{6} + \frac{6t^4}{4} - \frac{8t^3}{3} + t^2 \right) \Bigg|_0^2 .\\ \amp = \frac{64}{3} + 24 - \frac{64}{3} + 4 = 28 \end{align*}

Activity 10.3.2.

Calculate the line integral of \(F = (x^2-y^2,x+y)\) over the parametric curve \(\gamma = (\cosh t, \sinh t)\) for \(t \in [0,\ln 4]\text{.}\)
Solution.
I need to evalute the field along the curve, by replacing the variables defining the field with the matching components of the curve.
\begin{align*} F(\gamma(t)) \amp = (\cosh^2 t - \sinh^2 t, \cosh t + \sinh t) = (1, \cosh t + \sinh t) \end{align*}
Next I need to calculate the tangent of the curve.
\begin{align*} \gamma^\prime(t) \amp = (\sinh t, \cosh t) \end{align*}
Then I need the dot product of the two previous calculations.
\begin{align*} F(\gamma(t)) \cdot \gamma^\prime(t) \amp = \sinh t + \cosh^2 t + \sinh t \cosh t \end{align*}
Finally, I do the line integral by integrating this dot product over the parameter domain.
\begin{align*} \amp \int_0^{\ln 4} F(\gamma(t)) \cdot \gamma^\prime(t) dt \\ \amp = \int_0^{\ln 4} \sinh t + \cosh^2 t + \sinh t \cosh t\\ \amp = \left( \cosh t + \frac{t}{2} + \frac{\sinh 2t }{4} + \frac{\cosh^2 t}{2} \right) \Bigg|_0^{\ln 4} \\ \amp = \frac{4 + \frac{1}{4}}{2} + \frac{\ln 4}{2} + \frac{16 - \frac{1}{16}}{4} + \frac{1}{2} \left( \frac{4 + \frac{1}{4}}{2} \right)^2 - 1 - 0 - 0 - 1 \\ \amp = \frac{\ln 4}{2} + \frac{17}{8} + \frac{255}{64} + \frac{289}{128} - 2 = \frac{128 \ln 2 + 1071}{128} \end{align*}

Activity 10.3.3.

Calculate the line integral of \(F = (y^2,x-z,x^2+yz)\) over the parametric curve \(\gamma = (3t,t^2-1,-4t)\) for \(t \in [0,5]\text{.}\)
Solution.
I need to evalute the field along the curve, by replacing the variables defining the field with the matching components of the curve.
\begin{align*} F(\gamma(t)) \amp = ((t^2-1)^2, 3t - (-4t), (3t)^2 - (t^2-1)(-4t)) \\ \amp = (t^4 - 2t^2 + 1, 7t, 9t^4 - 4t^3 + 4t^2) \end{align*}
Next I need to calculate the tangent of the curve.
\begin{align*} \gamma^\prime(t) \amp = (3, 2t, -4) \end{align*}
Then I need the dot product of the two previous calculations.
\begin{align*} F(\gamma(t)) \cdot \gamma^\prime(t) \amp = 3(t^4-2t^2+1) + (2t)(7t) - 4(9t^4 - 4t^3 + 4t^2)\\ \amp = 3t^4 - 6t^2 + 3 + 14t - 36t^4 + 16t^3 - 16t^2\\ \amp = -33t^4 + 16t^3 - 22t^2+ 14t + 3 \end{align*}
Finally, I do the line integral by integrating this dot product over the parameter domain.
\begin{align*} \int_0^{5} F(\gamma(t)) \cdot \gamma^\prime(t) dt \amp = \int_0^5 -33t^4 + 16t^3 - 22t^2 + 14t + 3 dt\\ \amp = \left( \frac{-33t^5}{5} + 4t^4 - \frac{22t^3}{3} + 7t^2 + 3t \right) \Bigg|_0^5\\ \amp = -20625 + 2500 - \frac{2750}{3} + 175 + 15 = \frac{-56555}{3} \end{align*}

Activity 10.3.4.

Calculate the line integral of \(F = (z + x, z + y, x^2 + y^2)\) over the parametric curve \(\gamma = \left( \cos t, \sin t, \frac{t}{2\pi} \right)\) for \(t \in [0,6\pi]\text{.}\)
Solution.
I need to evalute the field along the curve, by replacing the variables defining the field with the matching components of the curve.
\begin{align*} F(\gamma(t)) \amp = \left( \frac{t}{2\pi} + \cos t, \frac{t}{2\pi} + \sin t, 1 \right) \end{align*}
Next I need to calculate the tangent of the curve.
\begin{align*} \gamma^\prime(t) \amp = \left( -\sin t, \cos t, \frac{1}{2\pi} \right) \end{align*}
Then I need the dot product of the two previous calculations.
\begin{align*} F(\gamma(t)) \cdot \gamma^\prime(t) \amp = \frac{-t\sin t}{2\pi} - \sin t \cos t + \frac{t\cos t}{2\pi} + \sin t \cos t + \frac{1}{2\pi} \\ \amp = \frac{1}{2\pi} ( t \cos t - t \sin t + 1) \end{align*}
Finally, I do the line integral by integrating this dot product over the parameter domain.
\begin{align*} \int_0^{6\pi} F(\gamma(t)) \cdot \gamma^\prime(t) dt \amp = \int_0^{6\pi} \frac{1}{2\pi} (t \cos t - t \sin t + 1) dt \\ \amp = \frac{1}{2\pi} \left(t \sin t + \cos t - (\sin t - t \cos t) - t \right) \Bigg|_0^{6\pi}\\ \amp = \frac{1}{2\pi} (0 + 1 - (0 - 6\pi) - 6\pi - (0 + 1 - (0 - 0))\\ \amp = \frac{1}{2\pi} (1 - 12 \pi- 1 ) = -12 \end{align*}

Subsection 10.3.2 Activity

Activity 10.3.5.

Check that this vector field is conservative. (Make sure you restrict to a simply connected domain). Calculate the line integral over the given curve by finding a potential for the field and then evaluating the potential at the endpoint of the curve.
\begin{align*} F \amp = (yz,xz,xy)\\ \gamma(t) \amp = (t,t^2,t^3) \amp \amp t \in [0,3] \end{align*}
Solution.
The domain is all of \(\RR^3\text{,}\) which is simply connected. I check that the field is conservative on this domain by calculating the curl.
\begin{align*} \nabla \times F \amp = \left( \frac{d}{dy} xy - \frac{d}{dz} xz, \frac{d}{dz} yz - \frac{d}{dx} xy, \frac{d}{dx} xz - \frac{d}{dy} yz \right) \\ \amp = (x-x, y-y, z-z) = (0,0,0) \end{align*}
The curl is zero, so the field is conservative. Now I need to find a potential. I integrate each component of the field in the matching variable. I put together these three pieces to find a potential that matches all three.
\begin{align*} \int F_1 dx \amp = \int yz dz = xyz + c \\ \int F_2 dy \amp = \int xz dz = xyz + c \\ \int F_3 dz \amp = \int xy dz = xyz + c \\ f \amp = xyz \end{align*}
Next I find the endpoints of the curve and evaluate the potential at the endpoints. The different of these two values is the value of the line integral, by the fundamental theorem.
\begin{align*} f(\gamma(0)) \amp = f((0,0,0)) = 0 \\ f(\gamma(3)) \amp = f((3,9,27)) = 729 \\ f(\gamma(3)) - f(\gamma(0)) \amp = 729 - 0 = 729 \end{align*}

Activity 10.3.6.

Check that this vector field is conservative. (Make sure you restrict to a simply connected domain). Calculate the line integral by finding a potential for the field and then evaluating the potential at the endpoint of the curve.
\begin{align*} F \amp = (2,-3,-2)\\ \gamma(t) \amp = (4 \cos t, 4 \sin t, t^2) \amp \amp t \in [0,4\pi] \end{align*}
Solution.
The domain is all of \(\RR^3\text{,}\) which is simply connected. I check that the field is conservative on this domain by calculating the curl.
\begin{align*} \nabla \times F \amp = \left( \frac{d}{dy} -2 - \frac{d}{dz} -3, \frac{d}{dz} 2 - \frac{d}{dx} -2, \frac{d}{dx} -3 - \frac{d}{dy} 2 \right) \\ \amp = (0,0,0) \end{align*}
The curl is zero, so the field is conservative. Now I need to find a potential. I integrate each component of the field in the matching variable. I put together these three pieces to find a potential that matches all three.
\begin{align*} \int F_1 dx \amp = \int 2 dx = 2x + c\\ \int F_2 dy \amp = \int -3 dy = -3y + c\\ \int F_3 dz \amp = \int -2 dz = -2z + c\\ f \amp = 2x - 3y - 2z \end{align*}
Next I find the endpoints of the curve and evaluate the potential at the endpoints. The different of these two values is the value of the line integral, by the fundamental theorem.
\begin{align*} f(\gamma(0)) \amp = f((4,0,0)) = 8\\ f(\gamma(4\pi)) \amp = f((4,0,16\pi^2)) = 8 - 32\pi^2\\ f(\gamma(4\pi)) - f(\gamma(0)) \amp = -32\pi^2 \end{align*}

Activity 10.3.7.

Check that this vector field is conservative. (Make sure you restrict to a simply connected domain). Calculate the line integral by finding a potential for the field and then evaluating the potential at the endpoint of the curve.
\begin{align*} F \amp = (\ln x, \ln z, \frac{y}{z}) \\ \gamma(t) \amp = (4t,-t,t^2) \amp \amp t \in [1,5] \end{align*}
Solution.
The domain is the portion of \(\RR^3\) where \(x \gt 0\text{,}\) \(z \gt 0\text{.}\) This is a simply connected domain. Moreover, the output of the parametric curve always have positive \(x\) and \(z\) coordinates, so it lies entirely within this domain. I check that the field is conservative on this domain by calculating the curl.
\begin{align*} \nabla \times F \amp = \left( \frac{d}{dy} \frac{y}{z} - \frac{d}{dz} \ln z, \frac{d}{dz}\ln x - \frac{d}{dx} \frac{y}{z}, \frac{d}{dx} \ln z - \frac{d}{dy} \ln x \right)\\ \amp = \left( \frac{1}{z} - \frac{1}{z}, 0 - 0 , 0 - 0 \right) = (0,0,0) \end{align*}
The curl is zero, so the field is conservative. Now I need to find a potential. I integrate each component of the field in the matching variable. I put together these three pieces to find a potential that matches all three.
\begin{align*} \int F_1 dx \amp = \int \ln x dx = x \ln x - x + c\\ \int F_2 dy \amp = \int \ln z dy = y \ln z + c\\ \int F_3 dz \amp = \int \frac{y}{z} dz = y \ln z + c\\ f \amp = x \ln x - x + y \ln z \end{align*}
Next I find the endpoints of the curve and evaluate the potential at the endpoints. The different of these two values is the value of the line integral, by the fundamental theorem.
\begin{align*} f(\gamma(1)) \amp = f((4,-1,4)) = 20 \ln 20 - 20 - 5 \ln 25 \\ f(\gamma(5)) \amp = f((20, -5, 25)) = 4 \ln 4 - 4 - \ln 4 \\ f(\gamma(5)) - f(\gamma(1)) \amp = 20 \ln 20 - 5 \ln 25 - 3 \ln 4 - 16 \end{align*}

Activity 10.3.8.

Check that this vector field is conservative. (Make sure you restrict to a simply connected domain). Calculate the line integral by finding a potential for the field and then evaluating the potential at the endpoint of the curve.
\begin{align*} F \amp = \left( \sqrt{y^2-z^2}, \frac{yx}{\sqrt{y^2-z^2}}, \frac{-xz}{\sqrt{y^2-z^2}} \right)\\ \gamma(t) \amp = (t+1,-3t-4,-t-1) \amp \amp t \in [0,4] \end{align*}
Solution.
The domain restriction is \(y^2 \gt z^2\text{.}\) This can be expanded into \(y \gt |z|\) or \(y \lt -|z|\text{.}\) This is not a simply connected domain (not even connected at all). I need to chose a region that satisfy these inequality and also include the parametric curve. For the parametric curve, both \(y\) and \(z\) are negative, so I can take the region \(y \lt z \lt 0\text{.}\) This is a simply connected domain. I check that the field is conservative on this domain by calculating the curl.
\begin{align*} \nabla \times F \amp = \left( \frac{d}{dy} \frac{-xz}{\sqrt{y^2-z^2}} - \frac{d}{dz} \frac{yx}{\sqrt{y^2-z^2}}, \frac{d}{dz} \sqrt{y^2-z^2} - \frac{d}{dx} \frac{-xz}{\sqrt{y^2-z^2}}, \right.\\ \amp \left. \frac{d}{dx} \frac{yx}{\sqrt{y^2-z^2}} - \frac{d}{dy} \sqrt{y^2-z^2} \right)\\ \amp = \left( \frac{xyz}{\sqrt{(y^2-z^2)^3}} - \frac{xyz}{\sqrt{(y^2-z^2)^3}}, \frac{z}{\sqrt{y^2-z^2}} - \frac{z}{\sqrt{y^2-z^2}}, \right. \\ \amp \left. \frac{y}{\sqrt{y^2-z^2}} - \frac{y}{\sqrt{y^2-z^2}} \right) = (0,0,0) \end{align*}
The curl is zero, so the field is conservative. Now I need to find a potential. I integrate each component of the field in the matching variable. I put together these three pieces to find a potential that matches all three.
\begin{align*} \int F_1 dx \amp = \int \sqrt{y^2-z^2} dx = x\sqrt{y^2-z^2} + c\\ \int F_2 dy \amp = \int \frac{yx}{\sqrt{y^2-z^2}} = x\sqrt{y^2-z^2} + c\\ \int F_3 dz \amp = \int \frac{-xz}{\sqrt{y^2-z^2}} = x\sqrt{y^2-z^2} + c\\ f \amp = x\sqrt{y^2-z^2} \end{align*}
Next I find the endpoints of the curve and evaluate the potential at the endpoints. The different of these two values is the value of the line integral, by the fundamental theorem.
\begin{align*} f(\gamma(0)) \amp = f((1,-4,-1)) = \sqrt{15} \\ f(\gamma(4)) \amp = f((5, -16, -5)) = 5 \sqrt{231} \\ f(\gamma(4)) - f(\gamma(0)) \amp = 5 \sqrt{231} - \sqrt{15} \end{align*}

Subsection 10.3.3 Conceptual Review Questions

  • What is a line integral and what does it measure?
  • What is a conservative vector field?
  • What is the basic structure of the fundmanetal theorem of calculus and how is that structure extended into multivariable calculus?
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