Welcome to calculus. I'm professor Greist. We're about to begin lecture
31 on complex areas. >> Computing areas by means of a definite
integral is one of the classic applications of calculus. But it's not always so straight forward. Some shapes are too complex to be easily
decomposed into triangles or rectangles. This is especially true in
the context of applications where simplicity must yield to reality. In this lesson, we'll learn how to handle some of those
problems by using the area element. >> We begin this lesson with a quick, classical example of
computing the area between two curves, y-x = 0 and y squared + x = 2. The first of these is the diagonal line,
slope one. The second is a parabola
that opens to the left. We need to compute the area
of the region between them. I'll leave it to you to do
the substitution and algebra to determine the intersection points at one comma one,
and negative two comma, negative two. And now to compute the area,
we could proceed as before by sweeping out with a vertical
strip as an area element. But if we do this, then we need to decompose the region
into two different parts. The area element is, on the first part, as x goes from negative two to one,
the difference in height. We need to solve for the y coordinates. In doing so, we can write the area
element as the top y equals x minus negative square
root of two minus x. This height is multiplied by the width dx. For the right hand side of this
region as x is going from one to two. Then the area element is twice
square root of two minus x times dx. And to compute the area,
we need to integrate this area element. That means we're going to have to
write down a pair of integrals. The integral is x goes from negative
two to one of x plus root two minus x, dx plus the integral as x goes from one
to two of twice root two minus x, dx. Now neither of these is
a terribly difficult integral. But we're not going to do it. Why? Because there is a simpler method. We're going to reverse. The direction of integration and integrate with respect to y,
instead of with respect to x. And so, solving for the x coordinate and sweeping out via a horizontal
strip as our area element we see there is a uniform area element and
it will reduce to a single integral. What is this area element? If we take the x coordinate on the right,
that's two minus y squared. Subtract the x coordinate on the left,
namely y. Multiply by the thickness d-y. Then we can integrate this area element. As y goes from negative two to one. This is a much simpler integral, and we get, simply,
2y minus y cubed over three, minus y squared over two,
evaluated from negative two to one. That equals nine-halves. The area. The direction matters. If you encounter integral that is too
difficult, try reversing the direction. A related albeit more complex
example comes to us from thermodynamics in the analysis
of an idealized heat engine. What happens to a gas in a chamber,
in something that, say, might power an automobile? Now we're going to go over a couple of
terms from thermodynamics, very quickly. All of these have to do with measuring
what is happening to the volume, V, and the pressure, P,
of the gas in the chamber. In this idealized heat engine,
there are four stages, or strokes. The first is called a combustion stage, otherwise known as
an isothermal expansion. The volume is increasing and the pressure,
P, times the volume, V, is constant. Let's call that A. So in this first stroke,
you're moving to the right along a curve in the V-P plane when
a product is a constant. The second stroke is the power stroke. Sometimes called an isotropic expansion. In this case you're moving along the curve p times v to the gamma equals
another constant let's say B. Here, gamma is yet another constant
that is greater than one and depends upon the properties of the gas. The third stroke, or stage,
is the exhaust stroke. This is an isothermal contraction,
in which V is decreasing, and you're moving along a curve,
P times V equals another constant. Let's call this one a sub zero, and we'll
call the former PV constant a sub one. Now we’re moving to the left
along one of these curves. The last stroke,
is the compression stroke. Otherwise known as
an isentropic contraction. Volume is decreasing, and
you move along the curve. PV to the gamma equals a constant. We’ll call this fourth constant B naught, replacing the B in our earlier
isentropic expansion with B1. All right, that's a lot of setup. What's the payoff here? In thermodynamics, you trace out
this region in the V-P plane, and the work that the engine does
over four strokes of a cycle is equal to the area that is
traced out in this plane. And so we're motivated to find that area. If we sweep out this region,
integrating with respect to V, using vertical strips, it's not so simple. It seems to have several different pieces. Perhaps reversing
the direction would help? No, this does not make
things any simpler at all. Let's go back to integrating
with respect to V. In this case, the region breaks
up into three distinct zones. Each requiring its own integral. In the first piece,
we have on the top the curve PV equals A1. Solving for P gives us A1/V. Subtracting the P value for the bottom curve Means subtracting
B naught over V to the gamma. That's for the first region. The second region is between
two isothermal curves. That gives us A1 minus A naught over V. The third region,
with a little bit of work gives us an integrant of B1 over V to
the gamma minus A naught over V. Now, none of these integrals
is difficult to perform. What is difficult is determining the
limits of integration because we have to find all four intersection points of these
curves, and determine their v values. I'll leave it to you to show
with not a small amount of work, that the first intersection point
on the left is quantity b naught over a1 raised to the power,
one minus gamma. You can imagine how much work
it's going too take to do and simplify these integrals. That's the bad news. The good news is that the answer
at the end is really quite simple. A one, minus A not, over gamma, minus one,
times log of B one, over B naught. If the answer was that simple, well isn't
there an easier way to do the integral? And there is, but
you're going to have to wait to find out. Complicated shapes require new techniques. And here's one approach. In the case where a shape is
defined by a radial distance. We can use what is called
polar coordinates. So if we have a shape of
the form r equals f of theta, where r is a distance to the origin. Theta is an angle made to the X axis. Then, trying to fill out
such a region by vertical or horizontal strips as an area element
is likely to lead to frustration. And so, as we did in the case of
a circular disk, we can use an angular wedge, and
sweep that around based at the origin. What will this area element look like? We'll approximate it as a triangle,
whose length is r, the radial distance, and
whose height is r times d theta. In this case, the area element
is the area of this triangle, 1/2 r (r d theta). And that is 1/2 r squared d theta, but remember r is given as a function,
f of theta, hence, we have something that we
can integrate to get the area. That being, the integral of one-half
quantity f of theta squared, d theta. Well let's explore this in
the context of an example. Compute the area between the regions. R less than or equal to two sine theta,
and r greater than or equal to one. We can clearly see that r equals
one is a circle, radius one, centered at the origin. I'll leave it to you to discover that
r equals two sine theta is likewise a circle of radius one
shifted up the y axis. These curves intersect at two points. What are they? Well, by setting the r values equal and
solving for theta, we see that these points are at theta
equals pi over six and five pi over six. Now we want to compute the area
of the region between the one and the other by sweeping out
with a angular wedge. This area element is going
to look like the difference between two triangles
centered at the origin. The area element, dA in this case,
is going to be, what? Well, we take the outer area element that is one-half of
the outer radius squared and subtract off one-half times
the inner radius squared. Of course, the outer radius
is r equals two sin theta, the inner radius is r equals one, paying
careful attention to our inequalities. We have to perform this
integration as theta goes from pi over six to five pi over six. And so we see,
that with a little bit of simplification using the double angle formula,
we can reduce this integral to one minus cosine
two theta minus a half d theta. Simplifying a little bit,
we can integrate negative cosine two theta to minus
one half sin two theta, and one-half becomes one-half theta. Evaluating this from pi over six to
five pi over six, gives us, with work, our answer, pi over three plus
square root of three over two. Now, these examples, though complicated, illustrate a deeper truth, namely, that by changing coordinates,
we can make the integral simpler. It looks like we're playing around with
triangles and area elements like that. But what we're really doing
is changing coordinates, so that we can integrate
with respect to theta. Likewise in the problem that we
didn't solve about thermodynamics. The right way to solve that
problem is to change to a new coordinate system in which the associated
shape becomes a rectangle. That will make things easier to handle. You're gonna have to wait until
you get to multi-variable calculus to see how to do this properly. >> Through solving or at least attempting
some more challenging problems, we're led to a new perspective,
that of a change of variables formula for planar two-dimensional shapes. We're going to save the full exposition of
those ideas for multi-variable calculus. But if you'd like to see a preview,
check out the bonus material. Then, we'll move on to our next lesson, where we'll compute volumes.