### Video Transcript

Calculate the integral over πΆ of
π of π₯, π¦ dπ for the function π of π₯, π¦ and curve πΆ, where π of π₯, π¦ is
equal to π₯ plus π¦ squared and πΆ is the path from two, zero counterclockwise along
the circle π₯ squared plus π¦ 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
π‘ that uniquely defines every point on πΆ 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 π¦
equal two distinct functions of the same parameter π‘. So π₯ equals π of π‘ and π¦ equals
β of π‘.

Our integral of π of π₯, π¦ dπ
then becomes the integral of π of π of π‘, β of π‘ dπ . dπ is the line element
which is given by the square root of dπ₯ squared plus dπ¦ squared. And this can be rewritten as the
square root of dπ₯ by dπ‘ squared plus dπ¦ by dπ‘ squared times dπ‘. Our integral then becomes the
integral over πΆ of π of π of π‘, β of π‘ times the square root of π prime of π‘
squared plus β prime of π‘ squared dπ‘, where π prime of π‘ is dπ₯ by dπ‘ and β
prime of π‘ is dπ¦ by dπ‘.

Letβs 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 πΆ 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 π¦
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 πΆ into two separate curves, πΆ one and __ two, where πΆ
one is the semicircular region of the curve and πΆ 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 π‘.

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 πΆ one, we know that π₯ squared plus π¦ squared is equal to four. Substituting in the
parameterization for π₯, we get π‘ squared plus π¦ squared equals four. And rearranging for π¦ gives us π¦
equals the square root of four minus π‘ 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 π¦ will always be negative along this curve, which
clearly from the diagram it is not. We can get around this with our
knowledge that π¦ 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, π‘ had decreased
monotonically, but after this point, it doubles back and begins to increase again
with increasing π₯. So some values of π‘ will map to
multiple points on this curve. In this case, π¦ will be given by
both the positive and negative square root of four minus π‘ 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 πΆ 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 cos π‘ and
π¦ equal π sin π‘, where π is the distance of the curve from the origin and π‘ 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 couldnβt be
simpler. The radius of the semicircle is
completely constant, so it doesnβt even depend on the argument. The radius of this semicircle is
two, so π₯ equals two cos π‘ and π¦ equals two sin π‘. And we now have π₯ and π¦ uniquely
and concisely defined in terms of one parameter.

π‘ 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 πΆ one. Going back to our formula for the
line integral, we still need to find π prime of π‘ and β prime of π‘. In our case, we have simple
trigonometric derivatives, so π prime of π‘ is d by dπ‘ of two cos π‘, which is
just negative two sin π‘, and β prime of π‘ is d by dπ‘ of two sin π‘, which is just
two cos π‘. Substituting these into the
formula, we now have our integral over the curve πΆ one of π of two cos π‘, two sin
π‘ times the square root of negative two sin π‘ all squared plus two cos π‘ all
squared with respect to π‘. We can now reexpress our limits on
this integral in terms of π‘.

At the start of the curve πΆ one,
the argument with the π₯-axis is zero, so π‘ equals zero. And at the end of the curve πΆ one,
the argument with the π₯-axis is π, so π‘ equals π. So the limits of this integral
become π‘ equals zero and π‘ equals π. Now we need to find π of two cos
π‘, two sin π‘. π of π₯, π¦ is π₯ plus π¦
squared. And since π₯ equals two cos π‘ and
π¦ equals two sin π‘, then π of two cos π‘, two sin π‘ is two cos π‘ plus four sin
squared π‘. Letβs tidy this integrand up a
little. Expanding out the parentheses
inside this square root, we get four sin squared π‘ plus four cos squared π‘. Now using the trigonometric
identity, cos squared π‘ plus sin squared π‘ 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 π‘. In our case, we know that π‘ 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 of cos π‘ plus two sin squared π‘ with
respect to π‘. Letβs clear a little space and
proceed with the integration. The first term cos π‘ easily
integrates to sin π‘. Letβs 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 π‘ is
identically equal to one minus cos squared π‘. And from the double-angle formula,
cos squared π‘ is identically equal to one-half times one plus cos two π‘. Therefore, with some rearranging,
sin squared π‘ is identically equal to one-half times one minus cos two π‘. Our integral then becomes four
times the integral between zero and π of cos π‘ plus one minus cos two π‘ with
respect to π‘, which can now be easily integrated using recognition and the chain
rule. Integrating, we get four times sin
π‘ plus π‘ minus one-half sin two π‘ evaluated between zero and π. The sin of any multiple of π is
equal to zero. So when evaluated at zero and π,
sin π‘ and sin two π‘ will be equal to zero. Therefore, this evaluates to simply
four π.

This completes the line integral
for πΆ one, but we now need to find the line integral for the rest of the curve πΆ
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 πΆ one. If we let π₯ equal π‘, then π‘ will
monotonically increase from negative two, zero to two, zero. As for π¦, π¦ along this whole
section of the curve is just equal to zero. We then have π prime of π‘ is
equal to one and β prime of π‘ is equal to zero. Plugging this into our line
integral formula, we get to the integral over πΆ two of π of π‘, zero times the
square root of one squared plus zero squared with respect to π‘.

π‘ is just equal to π₯, and π₯ goes
from negative two to two. So the limits are just π‘ equals
negative two and π‘ equals two. π of π‘, zero is just π‘ plus zero
squared, and the square root becomes simply one. So this leaves us with simply the
integral of π‘ with respect to π‘. Integrating, we get one-half π‘
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 πΆ, πΆ two, is just equal to zero. Our complete line integral over the
curve πΆ is just the sum of the two separate line integrals πΌ πΆ one and πΌ πΆ
two. Therefore, the integral over πΆ of
π of π₯, π¦ with respect to π is equal to four π.