Chapter 1. calc_tutorial_17_2_008

Question 1.1

Recall Stokes' Theorem. Assume that S is an oriented surface with parametrization G: D → S, where D is a domain in the plane bounded by smooth, simple closed curves, and G is one-to-one and regular, except possibly at the boundary of D. More generally, S may be a finite union of surfaces of this type. Then Stokes' Theorem gives us the following equation.

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

The integral on the left is defined relative to the boundary orientation of ∂S. If S is closed (that is, ∂S is empty), then the surface integral on the right is v4/Yoxb7daTg0BC3/fg8y28QBiU=.

This equation also introduces the notation curl(F). Given a vector field F = <F₁, F₂, F₃> and letting the table below represent the 3 x 3 matrix A, we define curl(F) as follows.

curl(F) = det(A) = <(∂F₃ / ∂y) − (∂F₂ / ∂z), (∂F₁ / ∂z) − (∂F₃ / ∂x), (∂F₂ / ∂x) − (∂F₁ / ∂y)>

In order to verify Stokes' Theorem for the given problem, we must compute the integrals on both sides of the equation for Stokes' Theorem independently and show that they are equal. We can start by computing curl(F) to find the flux of the curl through the surface, represented by the right-hand side of the equation.

Find curl(F) for F = <y, x, x² + y²>.

curl(F) = <XvVM00l89Is=y, FmAhxBkQqN0=x, 0>

Question 1.2

To compute the flux of the curl through the surface, we will use the following Vector Surface Integral equation.

Let G(u, v) be an oriented parametrization of an oriented surface S with parameter domain D. Assume that G is one-to-one and regular, except possibly at points on the boundary of D. Then we calculate the vector surface integral using the following formula.

ZQZxGddQ4ujvw0mSiHGRGckXvxZqvlTxHWkhDGgnLPw1v61IRkt04qze1MmQ4TvjDP93UcMm5NL/M692LbnYN55lKNUB2JmDhqh7YTLR2s00qqP74r7kGno9lq9OPri9fYelGDV2tLQvEH/hsNMTSLaqB/vOszqjenYq0dopj/V7tbDB86p1v6NkhE4n1AXzj7DrOOrYW8ijjYjgP7BxlUl7yBQf9eLTu3SoZ26loMEFkpLJLZEBqUYZ5CvYDbXPvLaXsHUk7NsLbxL+xC7URPTmCr3qaNCRSHWez/ayVpAjj/Yq8nsX8dj9fiD8UDpWW9oi0emlunInZcLxYQPELeqdYce+kz+6S8PynbgODowUsSkTzAdPqUie6SIunXfEuNDwJAfHInOXZRXCnorKMXtIMegNRqExW9y+ILEMI75JePAMJeGhbU42yhq+BW9KRMe25/PfAWOCsa70agwyFu7Z3odEuKwxPnMeSH6a3q4j8oQ4NhTrD/NynvZtF4/x7zMr1JbfK3Uf/AxhTbq669a7BViRC6tyJOoo842ba4vTVRsBFtXA+0uw+EQg6xwfXuEM0DZ8A4AT/X6lYmmiHsq5NxgbtXU1IAiAMkGrTLUxxWVbriNdW0v2C/dN+cEFZTegKgTpVeRXebXB/8OzWiJ7JSZVIHMOTqdy+Mso2Gm9Yj3zikWsg19vK9r6vTwK5cB6bfS/2WJ17KUx0zvH7PaPtBL7ChHeGm3wS5JZTkqRZ81y9T7ciJIhxklMxrjYC/TIC3BRaVr93Y0QjAKAqPIk1iSbBIIPlaVhF7Uuh6uHChrqjFRf6AfR6cfFPM4EIpEmRN6gNa7vAGhLrFZD5C60iiTgyveTBjg5hXyS2l9q3eua4FktRARGz4WhdoUkXHSf2T8lA8z4b/DNdOQBGhCRnl53sMct9np0h25e7DKp6K2eSozz2lY7KFS4TRlkuJGN3+R7YoEh0+Mt7h8aTXpUXYgRi+OsGsB6Rxgw0QqJwlQSGyFMLsN/VOko4xi6kVCqY6UpRcI/Prsw4wlEuVPRiFALuCK1Wr25FfMdcq6X9bUl7Sp+2pvO5beg9ipfcCsYgp3PxTqVL415lF2rEvAFZCskbYw5ReJlH6vHrPqALDwWDk8V7i5jjJNRKcCY2Ou20LXq3j5dRRRRzlfPMlSxZ0lEK2Q/x1lIIarJRETlqCi0hyDPYtA34NRKFpTZvp9iExtdNRAxJLbk35bjxoYNU/B8NonY1t+c4pyKGkcBoE/WjO5+AokxgSIK/XFI/Z4CkntBDozfNNBsRZh0amYu2/uLZlwKEnAyTzEfrpYbaPixzQzahw5FZ1ujK0tm7Uhj147ol5CR4xLHvZCyViPQacfhnt2SYQzd7bU50tUTTBoodHjLciS+4mmyVkT9ShihfIfLzRK1b5BY/Whwy9ePKQkZ0hSDNquVg58AirB6GXQGJ3seNDud4jKrLGdFNrjXSs8/Pa5Y+wKR1wLykXVSy3hG8CEiO/xG2aKkgv3hziKXA5KU71W426qTlh/3dFeavvd7P4cPZLOJtw8FRkXMJNvMDxUT+pXGS+X69Aqinh0pT39F3SELQn8=

We wish to use this equation, but will replace F with the vector field curl(F).

In addition, we will need to parametrize the surface of the upper hemisphere of x² + y² + z² = $b. This is most easily done using 6cTxX6KT9ztS42rpfU4bJwbrfn0fKs3n6/LoOw== coordinates.

We use these standard coordinate conversions and parametrize the upper hemisphere as follows.

G(u, v) = <nc1ItEz0kR4=cos θ sin φ, nc1ItEz0kR4=sin θ sin φ, nc1ItEz0kR4=cos φ>

The bounds on θ and φ are 0 ≤ θ ≤

LL9psPjUb2HeWGS5YOn9xhW+jvHl1bbwN4ct4kPxoQ58E1J3bXtS86911P83jnkgkivzdDGRDUw3UhBQx3pu6X9uISw5mwbfpjl+Hxv2AFH5rHYmrnmhPX0rIrkivkvtc/a7xXn+b9Xbd7dwFb2u+jjAtN0FtXl1CRM35iv4EPd+jqVZPUtnznSybXhnc4f6SEpdE8D/4MpcdHTfto9NUz9/uZ/u5ywmgCv3WQZHNeOfbdpU0iNLKfCWmrDq34hSpXZUANBnLYDgiCgF3bgd6Y+KL4VTsFxol4GaS0Ft8g+ofWVmyAuwRJPLKtBgAMlr07cGorqXGJ69NOssMg34HVUcHQDRBTruhpTfMSTrl9mO/dCXPGblaR/p7Xb9GiV5TG9Rke9VcRFkwh6gv5TqSVW8aXWzBUb0uPyXnbqzGAQ+vbcWmM+O8dF1cqcr1LLJi4SMZNlQfvGYKBFi0/Mh5s+agu3CtM4HEIc4ePn/3orlh4pDzBtHIYvTCK5H/uL3OQ2rauJNQLkx/SMPdGfuZpp0rWN9JIwtoiqds/Chb0pWTiD6Z2Zecxwsdqea7QKC/teAqwUz1bvQ0s8t0ePy2yTSH5oe6FJmr5w1mk+WwnyYiPVrRCxXsMLjl8CUbJNWDC9oa04gXnP+RFBf9NuaVVahnHy0E+ymHcwM1hgsiwTg5U0kMRYMOwEJUT22v/KrEi0VYSt97H5Mxpz57z704GiIrOhiivkiBUQlIT+W4fQVN/6v

and

0 ≤ φ ≤

EAmBZbDk1ciR0EfqffctIZVSm9KRidnYlIcNGUJu1AIZWYdmZi7TlDUhpdxkUC0PWA3MNIEBMsBEQFDSsYMDZVAcr+XLX58ufm+q6OzDNIZiIP1Debo7YxqT7YR9xM0lfHOgQu/64Q4c8NMEtXr58QT1JDOw9Y9ce+VdJj6sApYKp8D8Lw992QMRWmw9OjMtyi5ZaCN+uRlpmgKy1utMdNU3Tfy8VzGIMX3YFm984vRoHgvKFijhj8YXw9qqkxcB31wFIReV6j11+H/Bej7yAhqiJzQZBoPlxJIETE6LVHpGOiZEPf0AEGjfI8v6VdJUuoFMDTVwGOc0OraoipKWTkTFXkM6RH0aPk8ikmHbM1jsCE21s9jcdkMm27d/5PU3n8yjlZHgJBlr/kbNM1/JbnglZdTh9k3U0oi6Ot8s2phOcRfnFwpqlCnDhcrLeVY5dJWyUqdmXrDAB55SYtOaY5dxbArWh8sFZUyPDeF1J3D+3xOYcEBhUROg/M93f8K52WMxwGq86bH5CWgagVE2PgmFtKO5OBbY3t13J4oqEcst2bMn0nc2MqCfGGTIKdtl8RKtK6bQLHqxqkper7HZ9rLqzmJamzg6N9mqtrs/EgbiDvgC3bjGv2TgzV2UHJLiGvznD+KueYt++11/Vfs/9xBPv8xrw07n1g6mYKVYyDSA0UPRBbCGWXCzV6KQcgjCejd457sCwoP8XX7vNsjNHGoJSncPiqLn3N5KrVsQLBKaCQ6r

With the notation in our problem, the appropriate vector surface integral that represents the flux of the curl through the surface is as follows.

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

Question 1.3

Use the parametrization of the hemisphere, G(u, v) = <$a cos θ sin φ, $a sin θ sin φ, $a cos φ> to compute n(φ, θ) = T_φ × T_θ the normal vector pointing outward through the top of the hemisphere.

T_φ = (∂G / ∂φ)

= <nc1ItEz0kR4=cos φ cos θ, nc1ItEz0kR4=cos φ sin θ, IHLqJALES1Y=sin φ>

T_θ = (∂G / ∂θ)

= <IHLqJALES1Y=sin φ sin θ, nc1ItEz0kR4=sin φ cos θ, 1Wh3cvJ2xF4=>

n(φ, θ) = T_φ × T_θ

= iSba6t70dtA=sin φ<sin φ cos θ, sin φ sin θ, cos φ>

Question 1.4

Use the parametrization of the hemisphere and the expression for curl(F) found in Step 1 to calculate curl(F(G(φ, θ))).

curl(F) = <2y, −2x, 0>

curl(F(G(φ, θ))) = <SFgqQUkJGdg=sin φ sin θ, YWSdEbNFjeo=sin φ cos θ, 1Wh3cvJ2xF4=>

Question 1.5

Compute the dot product curl(F(G(φ, θ))) · n(φ, θ).

curl(F(G(φ, θ))) · n(φ, θ) = <$c sin φ sin θ, -$c sin φ cos θ, 0> · $b sin φ<sin φ cos θ, sin φ sin θ, cos φ>

= 1Wh3cvJ2xF4=

Question 1.6

Compute the flux of the curl through the surface.

= 1Wh3cvJ2xF4=

Question 1.7

Now we must compute the line integral to verify Stokes' Theorem.

We will use the theorem for computing a vector line integral. If c(t) is a regular parametrization of an oriented curve C for a ≤ t ≤ b, then we have the following equation.

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

In this problem, the oriented curve C is the boundary of the hemisphere, ∂S. Since the surface S is the upper hemisphere with an upward-pointing normal vector, ∂S is the circle with radius $a in the xy-plane oriented in the x0bXiHjgUISwLIWzVHSleaGEE1fiGqYkI7O/SCSJWo4= direction. We can therefore parametrize the boundary as follows.

c(t) = <nc1ItEz0kR4=cos t, nc1ItEz0kR4=sin t, 1Wh3cvJ2xF4=>

The bounds on t are 0 ≤ t ≤

LL9psPjUb2HeWGS5YOn9xhW+jvHl1bbwN4ct4kPxoQ58E1J3bXtS86911P83jnkgkivzdDGRDUw3UhBQx3pu6X9uISw5mwbfpjl+Hxv2AFH5rHYmrnmhPX0rIrkivkvtc/a7xXn+b9Xbd7dwFb2u+jjAtN0FtXl1CRM35iv4EPd+jqVZPUtnznSybXhnc4f6SEpdE8D/4MpcdHTfto9NUz9/uZ/u5ywmgCv3WQZHNeOfbdpU0iNLKfCWmrDq34hSpXZUANBnLYDgiCgF3bgd6Y+KL4VTsFxol4GaS0Ft8g+ofWVmyAuwRJPLKtBgAMlr07cGorqXGJ69NOssMg34HVUcHQDRBTruhpTfMSTrl9mO/dCXPGblaR/p7Xb9GiV5TG9Rke9VcRFkwh6gv5TqSVW8aXWzBUb0uPyXnbqzGAQ+vbcWmM+O8dF1cqcr1LLJi4SMZNlQfvGYKBFi0/Mh5s+agu3CtM4HEIc4ePn/3orlh4pDzBtHIYvTCK5H/uL3OQ2rauJNQLkx/SMPdGfuZpp0rWN9JIwtoiqds/Chb0pWTiD6Z2Zecxwsdqea7QKC/teAqwUz1bvQ0s8t0ePy2yTSH5oe6FJmr5w1mk+WwnyYiPVrRCxXsMLjl8CUbJNWDC9oa04gXnP+RFBf9NuaVVahnHy0E+ymHcwM1hgsiwTg5U0kMRYMOwEJUT22v/KrEi0VYSt97H5Mxpz57z704GiIrOhiivkiBUQlIT+W4fQVN/6v

Question 1.8

Compute the dot product F(c(t)) · c'(t). Recall that F = <y, x,x² + y²> and c(t) = <$a cos t, $a sin t, 0>.

F(c(t)) = <nc1ItEz0kR4=sin t, nc1ItEz0kR4=cos t, iSba6t70dtA=>

c'(t) = <IHLqJALES1Y=sin t, nc1ItEz0kR4=cos t, 1Wh3cvJ2xF4=>

F(c(t)) · c'(t) = iSba6t70dtA=(cos²t - sin²t)

Question 1.9

Compute the line integral . Use the trigonometric identity cos 2t = cos²t − sin²t.

= U2GIbglD1oM=sin(2t)

= 1Wh3cvJ2xF4=

Therefore we have shown that , and we have verified Stokes' Theorem.