0:09

>> Computing areas by means of a definite integral is one of the classic

Â applications of calculus.

Â But it's not always so straight forward.

Â Some shapes are too complex to be easily decomposed into triangles or rectangles.

Â This is especially true in the context of applications

Â where simplicity must yield to reality.

Â In this lesson,

Â we'll learn how to handle some of those problems by using the area element.

Â >> We begin this lesson with a quick,

Â classical example of computing the area between

Â two curves, y-x = 0 and y squared + x = 2.

Â The first of these is the diagonal line, slope one.

Â The second is a parabola that opens to the left.

Â We need to compute the area of the region between them.

Â I'll leave it to you to do the substitution and algebra to determine

Â the intersection points at one comma one, and negative two comma, negative two.

Â And now to compute the area, we could proceed as before by

Â sweeping out with a vertical strip as an area element.

Â But if we do this,

Â then we need to decompose the region into two different parts.

Â 1:39

The area element is, on the first part,

Â as x goes from negative two to one, the difference in height.

Â We need to solve for the y coordinates.

Â In doing so, we can write the area element as the top y equals

Â x minus negative square root of two minus x.

Â This height is multiplied by the width dx.

Â For the right hand side of this region as x is going from one to two.

Â Then the area element is twice square root of two minus x times dx.

Â And to compute the area, we need to integrate this area element.

Â That means we're going to have to write down a pair of integrals.

Â The integral is x goes from negative two to one of x plus root two minus x,

Â dx plus the integral as x goes from one to two of twice root two minus x, dx.

Â Now neither of these is a terribly difficult integral.

Â But we're not going to do it.

Â Why?

Â Because there is a simpler method.

Â We're going to reverse.

Â The direction of integration and

Â integrate with respect to y, instead of with respect to x.

Â And so, solving for the x coordinate and

Â sweeping out via a horizontal strip as our area element

Â we see there is a uniform area element and it will reduce to a single integral.

Â What is this area element?

Â If we take the x coordinate on the right, that's two minus y squared.

Â Subtract the x coordinate on the left, namely y.

Â Multiply by the thickness d-y.

Â Then we can integrate this area element.

Â As y goes from negative two to one.

Â This is a much simpler integral, and

Â we get, simply, 2y minus y cubed over three,

Â minus y squared over two, evaluated from negative two to one.

Â That equals nine-halves.

Â The area.

Â The direction matters.

Â If you encounter integral that is too difficult, try reversing the direction.

Â 4:17

A related albeit more complex example comes to us from

Â thermodynamics in the analysis of an idealized heat engine.

Â What happens to a gas in a chamber, in something that, say,

Â might power an automobile?

Â Now we're going to go over a couple of terms from thermodynamics, very quickly.

Â All of these have to do with measuring what is happening to the volume, V, and

Â the pressure, P, of the gas in the chamber.

Â In this idealized heat engine, there are four stages, or strokes.

Â The first is called a combustion stage,

Â otherwise known as an isothermal expansion.

Â The volume is increasing and the pressure, P, times the volume, V, is constant.

Â Let's call that A.

Â So in this first stroke, you're moving to the right along

Â a curve in the V-P plane when a product is a constant.

Â The second stroke is the power stroke.

Â Sometimes called an isotropic expansion.

Â In this case you're moving along the curve

Â p times v to the gamma equals another constant let's say B.

Â Here, gamma is yet another constant that is greater than one and

Â depends upon the properties of the gas.

Â The third stroke, or stage, is the exhaust stroke.

Â This is an isothermal contraction, in which V is decreasing,

Â and you're moving along a curve, P times V equals another constant.

Â Let's call this one a sub zero, and we'll call the former PV constant a sub one.

Â 6:19

Now weâ€™re moving to the left along one of these curves.

Â The last stroke, is the compression stroke.

Â Otherwise known as an isentropic contraction.

Â Volume is decreasing, and you move along the curve.

Â PV to the gamma equals a constant.

Â Weâ€™ll call this fourth constant B naught,

Â replacing the B in our earlier isentropic expansion with B1.

Â All right, that's a lot of setup.

Â What's the payoff here?

Â In thermodynamics, you trace out this region in the V-P plane,

Â and the work that the engine does over four strokes of a cycle

Â is equal to the area that is traced out in this plane.

Â And so we're motivated to find that area.

Â 7:18

If we sweep out this region, integrating with respect to V,

Â using vertical strips, it's not so simple.

Â It seems to have several different pieces.

Â Perhaps reversing the direction would help?

Â No, this does not make things any simpler at all.

Â Let's go back to integrating with respect to V.

Â In this case, the region breaks up into three distinct zones.

Â Each requiring its own integral.

Â In the first piece, we have on the top the curve PV equals A1.

Â 8:01

Solving for P gives us A1/V.

Â Subtracting the P value for

Â the bottom curve Means subtracting B naught over V to the gamma.

Â That's for the first region.

Â The second region is between two isothermal curves.

Â That gives us A1 minus A naught over V.

Â The third region, with a little bit of work gives us

Â an integrant of B1 over V to the gamma minus A naught over V.

Â 8:38

Now, none of these integrals is difficult to perform.

Â What is difficult is determining the limits of integration because we have to

Â find all four intersection points of these curves, and determine their v values.

Â I'll leave it to you to show with not a small amount of work,

Â that the first intersection point on the left is quantity b naught

Â over a1 raised to the power, one minus gamma.

Â You can imagine how much work it's going too take to do and

Â simplify these integrals.

Â That's the bad news.

Â The good news is that the answer at the end is really quite simple.

Â A one, minus A not, over gamma, minus one, times log of B one, over B naught.

Â If the answer was that simple, well isn't there an easier way to do the integral?

Â And there is, but you're going to have to wait to find out.

Â 9:43

Complicated shapes require new techniques.

Â And here's one approach.

Â In the case where a shape is defined by a radial distance.

Â We can use what is called polar coordinates.

Â So if we have a shape of the form r equals f of theta,

Â where r is a distance to the origin.

Â Theta is an angle made to the X axis.

Â Then, trying to fill out such a region by vertical or

Â horizontal strips as an area element is likely to lead to frustration.

Â And so, as we did in the case of a circular disk, we can use an angular

Â wedge, and sweep that around based at the origin.

Â What will this area element look like?

Â We'll approximate it as a triangle, whose length is r,

Â the radial distance, and whose height is r times d theta.

Â In this case, the area element is the area of this triangle,

Â 1/2 r (r d theta).

Â And that is 1/2 r squared d theta, but

Â remember r is given as a function, f of theta,

Â hence, we have something that we can integrate to get the area.

Â That being, the integral of one-half quantity f of theta squared, d theta.

Â 11:25

Well let's explore this in the context of an example.

Â Compute the area between the regions.

Â R less than or equal to two sine theta, and r greater than or equal to one.

Â We can clearly see that r equals one is a circle, radius one,

Â centered at the origin.

Â I'll leave it to you to discover that r equals two sine theta is likewise

Â a circle of radius one shifted up the y axis.

Â These curves intersect at two points.

Â What are they?

Â Well, by setting the r values equal and solving for theta,

Â we see that these points are at theta equals pi over six and five pi over six.

Â Now we want to compute the area of the region between the one and

Â the other by sweeping out with a angular wedge.

Â This area element is going to look like the difference

Â between two triangles centered at the origin.

Â 12:36

The area element, dA in this case, is going to be, what?

Â Well, we take the outer area element

Â that is one-half of the outer radius squared and

Â subtract off one-half times the inner radius squared.

Â Of course, the outer radius is r equals two sin theta,

Â the inner radius is r equals one, paying careful attention to our inequalities.

Â We have to perform this integration as theta goes from pi

Â over six to five pi over six.

Â And so we see, that with a little bit of simplification

Â using the double angle formula, we can reduce this

Â integral to one minus cosine two theta minus a half d theta.

Â Simplifying a little bit, we can integrate negative

Â cosine two theta to minus one half sin two theta, and

Â one-half becomes one-half theta.

Â Evaluating this from pi over six to five pi over six, gives us, with work,

Â our answer, pi over three plus square root of three over two.

Â 13:57

Now, these examples, though complicated,

Â illustrate a deeper truth, namely,

Â that by changing coordinates, we can make the integral simpler.

Â It looks like we're playing around with triangles and area elements like that.

Â But what we're really doing is changing coordinates, so

Â that we can integrate with respect to theta.

Â Likewise in the problem that we didn't solve about thermodynamics.

Â The right way to solve that problem is to change to a new

Â coordinate system in which the associated shape becomes a rectangle.

Â That will make things easier to handle.

Â You're gonna have to wait until you get to multi-variable calculus

Â to see how to do this properly.

Â 14:50

>> Through solving or at least attempting some more challenging problems,

Â we're led to a new perspective, that of a change of variables formula for

Â planar two-dimensional shapes.

Â We're going to save the full exposition of those ideas for multi-variable calculus.

Â But if you'd like to see a preview, check out the bonus material.

Â Then, we'll move on to our next lesson,

Â where we'll compute volumes.

Â