Video: Evaluating the Line Integral of a Function of Two Variables around a Semi-circular Path

Calculate ∫_(𝐢) 𝑓(π‘₯, 𝑦) d𝑠 for the function 𝑓(π‘₯, 𝑦) and curve 𝐢, where 𝑓(π‘₯, 𝑦) = π‘₯ + 𝑦² and 𝐢 is the path from (2, 0) counterclockwise along the circle π‘₯Β² + 𝑦² = 4 to the point (βˆ’2, 0) and then back to (2, 0) along the π‘₯-axis.

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Video Transcript

Calculate the integral over 𝐢 of 𝑓 of π‘₯, 𝑦 d𝑠 for the function 𝑓 of π‘₯, 𝑦 and curve 𝐢, where 𝑓 of π‘₯, 𝑦 is equal to π‘₯ plus 𝑦 squared and 𝐢 is the path from two, zero counterclockwise along the circle π‘₯ squared plus 𝑦 squared equals four to the point negative two, zero and then back to two, zero along the π‘₯-axis.

Recall that to find the line integral of a function of multiple variables along a curve, we need to define a bijective or one-to-one parameterization for the set of variables to some parameter 𝑑 that uniquely defines every point on 𝐢 and then integrate with respect to this parameter. What this means in practice in our case for a function of two variables is that we let our two variables π‘₯ and 𝑦 equal two distinct functions of the same parameter 𝑑. So π‘₯ equals 𝑔 of 𝑑 and 𝑦 equals β„Ž of 𝑑.

Our integral of 𝑓 of π‘₯, 𝑦 d𝑠 then becomes the integral of 𝑓 of 𝑔 of 𝑑, β„Ž of 𝑑 d𝑠. d𝑠 is the line element which is given by the square root of dπ‘₯ squared plus d𝑦 squared. And this can be rewritten as the square root of dπ‘₯ by d𝑑 squared plus d𝑦 by d𝑑 squared times d𝑑. Our integral then becomes the integral over 𝐢 of 𝑓 of 𝑔 of 𝑑, β„Ž of 𝑑 times the square root of 𝑔 prime of 𝑑 squared plus β„Ž prime of 𝑑 squared d𝑑, where 𝑔 prime of 𝑑 is dπ‘₯ by d𝑑 and β„Ž prime of 𝑑 is d𝑦 by d𝑑.

Let’s clear some space and move this up here out of the way. In this question, we are asked to find the line integral of a function of two variables without being given a parameterization for those variables. So we will need to introduce one of our own. The curve 𝐢 that we are asked to integrate over looks something like this. We go around a semicircle of radius two anticlockwise from the point two, zero to the point negative two, zero and then straight back along the π‘₯-axis from negative two, zero to two, zero. Any parameterization for π‘₯ and 𝑦 is valid so long as the function is bijective or one to one. What this means in practice for integrating a continuous function like this one is that the parameter must either monotonically increase or decrease as we traverse the curve from its start point to its end point.

Of course, we do not have to use the same parameterization for every part of the curve. In this case, we will find it much easier to split the curve 𝐢 into two separate curves, 𝐢 one and __ two, where 𝐢 one is the semicircular region of the curve and 𝐢 two is the straight region of the curve along the π‘₯-axis. The easiest way to parameterize a function of multiple variables is simply to let the first variable, in this case π‘₯, be equal to the parameter 𝑑.

We can then parameterize the other variables in the function by substituting in the parameterization for π‘₯. So along the first part of the curve 𝐢 one, we know that π‘₯ squared plus 𝑦 squared is equal to four. Substituting in the parameterization for π‘₯, we get 𝑑 squared plus 𝑦 squared equals four. And rearranging for 𝑦 gives us 𝑦 equals the square root of four minus 𝑑 squared, but we already have a problem. This is not a bijective parameterization for the curve, since the square root function is an injective or one-to-many function.

If, for instance, we take the negative square root here, then 𝑦 will always be negative along this curve, which clearly from the diagram it is not. We can get around this with our knowledge that 𝑦 is always positive along this curve and simply force the positive square root. However, this is not a particularly neat solution and only works because this specific curve stays above the π‘₯-axis.

Imagine, for instance, if this curve had continued past this point. Before this point, 𝑑 had decreased monotonically, but after this point, it doubles back and begins to increase again with increasing π‘₯. So some values of 𝑑 will map to multiple points on this curve. In this case, 𝑦 will be given by both the positive and negative square root of four minus 𝑑 squared. What is more even with the simplification that this specific curve remains above the π‘₯-axis, the integral that results from this parameterization is still extremely nasty. Try it for yourself if you dare.

The upshot of this is that when looking for a parameterization, some parameterizations will make the job much easier than others. In this case, for 𝐢 one, we have the arc of a circle, so there is one parameterization in particular that leaps out as the most natural, polar. We can let π‘₯ equal π‘Ÿ cos 𝑑 and 𝑦 equal π‘Ÿ sin 𝑑, where π‘Ÿ is the distance of the curve from the origin and 𝑑 is the argument that this point on the curve makes with the positive π‘₯-axis.

The polar parameterization is particularly convenient for any curve where the radius is easily defined by the argument because we can then in turn express the radius in terms of the argument so that we have only one parameter. In this case, it couldn’t be simpler. The radius of the semicircle is completely constant, so it doesn’t even depend on the argument. The radius of this semicircle is two, so π‘₯ equals two cos 𝑑 and 𝑦 equals two sin 𝑑. And we now have π‘₯ and 𝑦 uniquely and concisely defined in terms of one parameter.

𝑑 will, of course, increase monotonically from the point two, zero to the point negative two, zero. So it will map uniquely to exactly one point along the curve 𝐢 one. Going back to our formula for the line integral, we still need to find 𝑔 prime of 𝑑 and β„Ž prime of 𝑑. In our case, we have simple trigonometric derivatives, so 𝑔 prime of 𝑑 is d by d𝑑 of two cos 𝑑, which is just negative two sin 𝑑, and β„Ž prime of 𝑑 is d by d𝑑 of two sin 𝑑, which is just two cos 𝑑. Substituting these into the formula, we now have our integral over the curve 𝐢 one of 𝑓 of two cos 𝑑, two sin 𝑑 times the square root of negative two sin 𝑑 all squared plus two cos 𝑑 all squared with respect to 𝑑. We can now reexpress our limits on this integral in terms of 𝑑.

At the start of the curve 𝐢 one, the argument with the π‘₯-axis is zero, so 𝑑 equals zero. And at the end of the curve 𝐢 one, the argument with the π‘₯-axis is πœ‹, so 𝑑 equals πœ‹. So the limits of this integral become 𝑑 equals zero and 𝑑 equals πœ‹. Now we need to find 𝑓 of two cos 𝑑, two sin 𝑑. 𝑓 of π‘₯, 𝑦 is π‘₯ plus 𝑦 squared. And since π‘₯ equals two cos 𝑑 and 𝑦 equals two sin 𝑑, then 𝑓 of two cos 𝑑, two sin 𝑑 is two cos 𝑑 plus four sin squared 𝑑. Let’s tidy this integrand up a little. Expanding out the parentheses inside this square root, we get four sin squared 𝑑 plus four cos squared 𝑑. Now using the trigonometric identity, cos squared 𝑑 plus sin squared 𝑑 is identically equal to one, this whole expression is just equal to four.

We now take the positive square root of four since the positive or negative direction is given by the increase or decrease of 𝑑. In our case, we know that 𝑑 is increasing from the point two, zero to the point negative two, zero, so we take the positive square root. This is a constant, so we can take it outside the integration along with a factor of two from the first set of parentheses. So this whole integral becomes four times the integral between zero and πœ‹ of cos 𝑑 plus two sin squared 𝑑 with respect to 𝑑. Let’s clear a little space and proceed with the integration. The first term cos 𝑑 easily integrates to sin 𝑑. Let’s just mark this here for now and look at the second part of the integral.

For the second term, we will need to make some use of some trigonometric identities. Firstly, sin squared 𝑑 is identically equal to one minus cos squared 𝑑. And from the double-angle formula, cos squared 𝑑 is identically equal to one-half times one plus cos two 𝑑. Therefore, with some rearranging, sin squared 𝑑 is identically equal to one-half times one minus cos two 𝑑. Our integral then becomes four times the integral between zero and πœ‹ of cos 𝑑 plus one minus cos two 𝑑 with respect to 𝑑, which can now be easily integrated using recognition and the chain rule. Integrating, we get four times sin 𝑑 plus 𝑑 minus one-half sin two 𝑑 evaluated between zero and πœ‹. The sin of any multiple of πœ‹ is equal to zero. So when evaluated at zero and πœ‹, sin 𝑑 and sin two 𝑑 will be equal to zero. Therefore, this evaluates to simply four πœ‹.

This completes the line integral for 𝐢 one, but we now need to find the line integral for the rest of the curve 𝐢 two. This also means we need to define a new parameterization. Fortunately, this second integral is much easier. And we can in fact use the parameterization we tried first before for 𝐢 one. If we let π‘₯ equal 𝑑, then 𝑑 will monotonically increase from negative two, zero to two, zero. As for 𝑦, 𝑦 along this whole section of the curve is just equal to zero. We then have 𝑔 prime of 𝑑 is equal to one and β„Ž prime of 𝑑 is equal to zero. Plugging this into our line integral formula, we get to the integral over 𝐢 two of 𝑓 of 𝑑, zero times the square root of one squared plus zero squared with respect to 𝑑.

𝑑 is just equal to π‘₯, and π‘₯ goes from negative two to two. So the limits are just 𝑑 equals negative two and 𝑑 equals two. 𝑓 of 𝑑, zero is just 𝑑 plus zero squared, and the square root becomes simply one. So this leaves us with simply the integral of 𝑑 with respect to 𝑑. Integrating, we get one-half 𝑑 squared evaluated between negative two and two. And this evaluates to one-half times two squared minus negative two squared, which is just two squared. So this whole thing is just equal to zero. Therefore, the line integral over the second part of the curve 𝐢, 𝐢 two, is just equal to zero. Our complete line integral over the curve 𝐢 is just the sum of the two separate line integrals 𝐼 𝐢 one and 𝐼 𝐢 two. Therefore, the integral over 𝐢 of 𝑓 of π‘₯, 𝑦 with respect to 𝑠 is equal to four πœ‹.

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