In order to use the fundamental theorem of line integrals, you need to find a scalar potential function - that is, a scalar function <em>f(x, y)</em> for which
grad <em>f(x, y)</em> = <em>F</em><em>(x, y)</em>
This amounts to solving for <em>f</em> such that
∂<em>f</em>/d<em>x</em> = <em>x</em> + 3
∂<em>f</em>/∂<em>y</em> = 6<em>y</em> + 3
Integrating both sides of the first equation with respect to <em>x</em> gives
<em>f</em> = 1/2 <em>x</em> ^2 + 3<em>x</em> + <em>g(y)</em>
Differentiating with respect to <em>y</em> gives
∂<em>f</em>/∂<em>y</em> = d<em>g</em>/d<em>y</em> = 6<em>y</em> + 3
Solving for <em>g</em> gives
<em>g</em> = ∫ (6<em>y</em> + 3) d<em>y</em> = 3<em>y</em> ^2 + 3<em>y</em> + <em>C</em>
and hence
<em>f(x, y)</em> = 1/2 <em>x</em> ^2 + 3<em>x</em> + 3<em>y</em> ^2 + 3<em>y</em> + <em>C</em>
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(a) By the fundamental theorem, the integral of <em>F</em> along any path starting at the point <em>P</em> (1, 0) and ending at <em>Q</em> (3, 3) is
∫ <em>F</em><em>(x, y)</em> • d<em>r</em> = <em>f</em> (3, 3) - <em>f</em> (1, 0) = 99/2 - 7/2 = 46
(b) Now we're talking about a closed path, so the integral is simply 0. We can verify this by checking the integral over the origin-containing paths:
• From the origin to <em>P</em> :
∫ <em>F</em><em>(x, y)</em> • d<em>r</em> = <em>f</em> (1, 0) - <em>f</em> (0, 0) = 7/2 - 0 = 7/2
• From <em>Q</em> back to the origin:
∫ <em>F</em><em>(x, y)</em> • d<em>r</em> = <em>f</em> (0, 0) - <em>f</em> (3, 3) = 0 - 99/2 = -99/2
Then the total integral is 7/2 + 46 - 99/2 = 0, as expected.
