Line Integrals - Part II — Question 7

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Question 7

Let CC be the vertical segment x=2x=2 from (2,0)(2,0) to (2,3)(2,3). Evaluate both ∫C(x+y)dxand∫C(x+y)dy.\int_C(x+y)\,dx \qquad\text{and}\qquad \int_C(x+y)\,dy.

Tasks

  1. Parametrize the vertical segment.

  2. Evaluate both integrals.

  3. Explain why treating yy as a function of xx is inappropriate here.

Original worksheet page 1: question and worked solution for 5-3-007
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Question 7 – Solution

Strategy. Parametrize by the changing coordinate yy; the constant coordinate has zero differential.

Step 1: Parametrize 𝒓(t)=⟨2,t⟩,0≤t≤3.\mathbf r(t)=\langle 2,t\rangle,\qquad 0\le t\le 3. Thus dx=0dx=0, dy=dtdy=dt, and x+y=2+tx+y=2+t.

See the diagram in the original worksheet below.

Step 2: Evaluate ∫C(x+y)dx=0\boxed{\int_C(x+y)\,dx=0} because dx=0dx=0 everywhere. Meanwhile, ∫C(x+y)dy=∫03(2+t)dt=[2t+t22]03=212.\int_C(x+y)\,dy =\int_0^3(2+t)\,dt =\left[2t+\frac{t^2}{2}\right]_0^3 =\boxed{\frac{21}{2}}.

Step 3: Representation issue A vertical segment assigns many yy-values to the single value x=2x=2, so it cannot be written as a single-valued graph y=g(x)y=g(x). Parametrization avoids this limitation.

Verification The second integrand rises linearly from 22 to 55; its average 7/27/2 times the yy-interval length 33 gives 21/221/2.

Original worksheet page 2: question and worked solution for 5-3-007

Original worksheet layout. Use Enlarge or open the PDF for a closer view.