Worksheet: Line Integrals of Vector Fields

In this worksheet, we will practice finding the line integral of a vector field along a curve with an orientation.

Q1:

Suppose 𝐢 is the path given by r(𝑑)=(𝑑,𝑑) for 0≀𝑑≀1, 𝐢 is the path given by r(𝑑)=(1βˆ’π‘‘,1βˆ’π‘‘) for 0≀𝑑≀1, and Fij=π‘₯+(𝑦+1)ln. Without calculating the integrals, which of the following is true?

  • Aο„Έβ‹…>ο„Έβ‹…οŒ’οŒ’οŽ οŽ‘FrFrdd
  • Bο„Έβ‹…=ο„Έβ‹…οŒ’οŒ’οŽ οŽ‘FrFrdd
  • Cο„Έβ‹…<ο„Έβ‹…οŒ’οŒ’οŽ οŽ‘FrFrdd

Q2:

In the figure, the curve 𝐢 from 𝑃 to 𝑄 consists of two quarter-unit circles, one with center (1, 0) and the other with center (3, 0). Calculate the line integral ο„Έβ‹…οŒ’Frd, where F=ο€»π‘₯2𝑦2sinsinijβˆ’ο€»π‘₯2𝑦2coscos.

  • A2ο€Ό12οˆο€Ό32+2ο€Ό12οˆο€Ό12sincoscossin
  • Bβˆ’2ο€Ό12οˆο€Ό12οˆβˆ’2ο€Ό12οˆο€Ό12sincoscossin
  • Cβˆ’2ο€Ό12οˆο€Ό32οˆβˆ’2ο€Ό32οˆο€Ό12sincoscossin
  • Dβˆ’2ο€Ό12οˆο€Ό12+2ο€Ό12οˆο€Ό12sincoscossin
  • E2ο€Ό12οˆο€Ό32+2ο€Ό12οˆο€Ό12sincoscossin

Q3:

Let 𝑃 be the arc of a unit circle in the π‘₯𝑦-plane traversed counterclockwise from (0,1) to (1,0). Determine the exact value of the line integral of the vector field Fijk(π‘₯,𝑦,𝑧)=3π‘₯𝑒+2𝑦𝑧𝑒+π‘¦π‘’οŠ¨ο—οŠ°ο˜ο™ο—οŠ°ο˜ο™οŠ¨ο—οŠ°ο˜ο™οŽ’οŽ‘οŽ’οŽ‘οŽ’οŽ‘ over 𝑃.

  • A1+𝑒
  • B1+2𝑒
  • C1βˆ’π‘’
  • Dπ‘’βˆ’1
  • E1βˆ’2𝑒

Q4:

We explore an example where a vector field F=⟨𝐹,𝐹⟩ satisfies πœ•πΉπœ•π‘¦βˆ’πœ•πΉπœ•π‘₯=0 but does not come from a potential. On the plane with the origin removed, consider the vector field F(π‘₯,𝑦)=ο“’βˆ’π‘¦π‘₯+𝑦,π‘₯π‘₯+π‘¦ο““οŠ¨οŠ¨οŠ¨οŠ¨.

On the (open) half-plane π‘₯>0, we can define the angle function πœƒ(π‘₯,𝑦)=𝑦π‘₯arctan. This is well defined and gives a value between βˆ’πœ‹2 and πœ‹2. What is the gradient βˆ‡πœƒ?

  • A𝑦π‘₯+𝑦,βˆ’π‘₯π‘₯+π‘¦ο““οŠ¨οŠ¨οŠ¨οŠ¨
  • Bο“’βˆ’π‘¦π‘₯+𝑦,π‘₯π‘₯+π‘¦ο““οŠ¨οŠ¨οŠ¨οŠ¨
  • C𝑦π‘₯+𝑦,π‘₯π‘₯+π‘¦ο““οŠ¨οŠ¨οŠ¨οŠ¨
  • Dο“’βˆ’π‘¦π‘₯+𝑦,βˆ’π‘₯π‘₯+π‘¦ο““οŠ¨οŠ¨οŠ¨οŠ¨
  • Eο“’βˆ’π‘₯π‘₯+𝑦,𝑦π‘₯+π‘¦ο““οŠ¨οŠ¨οŠ¨οŠ¨

Using the figure shown, use πœƒ above to define the angle function πœƒ(π‘₯,𝑦) on the region 𝑦>0 by a suitable composition with a πœ‹2 rotation.

  • Aπœƒ(π‘₯,𝑦)=πœƒ(βˆ’π‘¦,π‘₯)+πœ‹2
  • Bπœƒ(π‘₯,𝑦)=πœƒ(βˆ’π‘¦,π‘₯)
  • Cπœƒ(π‘₯,𝑦)=πœƒ(βˆ’π‘¦,βˆ’π‘₯)+πœ‹2
  • Dπœƒ(π‘₯,𝑦)=πœƒ(𝑦,βˆ’π‘₯)+πœ‹2
  • Eπœƒ(π‘₯,𝑦)=πœƒ(𝑦,βˆ’π‘₯)

What is βˆ‡πœƒ(π‘₯,𝑦)?

  • A𝑦π‘₯+𝑦,βˆ’π‘₯π‘₯+π‘¦ο““οŠ¨οŠ¨οŠ¨οŠ¨
  • B𝑦π‘₯+𝑦,π‘₯π‘₯+π‘¦ο““οŠ¨οŠ¨οŠ¨οŠ¨
  • Cο“’βˆ’π‘¦π‘₯+𝑦,π‘₯π‘₯+π‘¦ο““οŠ¨οŠ¨οŠ¨οŠ¨
  • Dο“’βˆ’π‘₯π‘₯+𝑦,𝑦π‘₯+π‘¦ο““οŠ¨οŠ¨οŠ¨οŠ¨
  • Eο“’βˆ’π‘¦π‘₯+𝑦,βˆ’π‘₯π‘₯+π‘¦ο““οŠ¨οŠ¨οŠ¨οŠ¨

Since πœƒ and πœƒοŠ§ agree on the quadrant π‘₯>0, 𝑦>0, we can define the angle 𝑇(π‘₯,𝑦) at each point of the union with values between βˆ’πœ‹2 and 3πœ‹2. Use the functions πœƒ and πœƒοŠ§ to find ο„Έβ‹…οŒ’Frd, where 𝐢 is the arc of the unit circle from ο€Ώ12,βˆ’βˆš32 to ο€Ώβˆ’1√2,1√2. Answer in terms of πœ‹.

  • A13πœ‹12
  • B3πœ‹5
  • Cπœ‹12
  • Dπœ‹5
  • E12πœ‹13

In the same way, we can define πœƒοŠ¨ on the half-plane π‘₯<0 and πœƒοŠ© on 𝑦<0. Hence, evaluate the line integral ο„Έβ‹…οŒ’Frd around the circle of radius √2, starting from 𝑃(1,βˆ’1) and going counterclockwise back to 𝑃, stating your answer in terms of πœ‹.

  • Aπœ‹2
  • Bπœ‹
  • Cπœ‹βˆš2
  • D√2πœ‹
  • E2πœ‹

Q5:

Suppose that F is the gradient of the function 𝑓(π‘₯,𝑦)=2π‘₯βˆ’π‘¦οŠ©οŠ¨ and we are given points 𝑃(0,0),𝑄(1,0),𝑅(0,1),𝑆(1,1), and 𝑇(βˆ’1,βˆ’1). Choose a starting and an end point from this set so as to maximize the integral ο„Έβ‹…οŒ’Frd, where 𝐢 is the line between your chosen points.

  • AFrom 𝑆 to 𝑄
  • BFrom 𝑅 to 𝑇
  • CFrom 𝑇 to 𝑄
  • DFrom 𝑃 to 𝑅
  • EFrom 𝑄 to 𝑇

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