 This lesson is on volumes using cross sections. When you went into the website, the author called this volumes by section. Most books, most teachers call this volumes using cross sections because we get an object and we cut it up on its cross sections like the loaf of bread that you saw in the website. So when you cut things up, you look at their cross sections or their areas. So let's go on and start our examples on these. And most of these examples have been taken from the problems on the website. Our first example is one from the website. It says the base is a unit square and the cross sections are squares with a height of four. And what you had seen was a pyramid, but I'm going to re-sketch that to look like this. And the reason I'm doing this is because I am taking my cross sections off of the particular axis and it will be perpendicular to the x-axis. So if I do that, each piece of my base is called y. And I have a point here because this is four units high. That point is four zero. And if I reach the end of my base, that is called four zero point five. So I have a formula for my line here and that's y is equal to one eighth x. And that's going to become important in a few minutes. So what are we doing? Well, we're slicing up this particular object into many, many squares. And if you just keep drawing those squares you see in this case they get smaller and smaller. So what we want to do is find the area of one of the squares and then using the Riemann sum idea add all those squares together to find the volume. So let's redraw that base. It looks like that. And then you notice I have a y here and a y going down. So really that base is 2y. If I find the area of that it's equal to 2y quantity squared. But because I am drawing this perpendicular to the x-axis all my areas need to be in x's. So I will transform my y to what it is in x's and it's two times. And if you see it's y is equal to one eighth x. It's one eighth x. And we're going to square that. Our y is equal to one fourth x quantity squared. So in doing that we can actually simplify a little bit more and get one over 16x squared. So that's the area of one of my squares. So to find the volume we are going to add all those areas up. So it's one over 16x squared. And of course there is a thickness, slight thickness on each one of these that we will use for our dx. And our interval will go from zero to four. In doing all that we will get a volume of one over 16x cubed over three. And we're going to evaluate it from zero to four. And we will get one over 16x64 over three. And our final answer will give us four thirds cubic units. So this is how you find the volume using cross sections. You take the information you have. You really draw a little sketch of that area to make sure that you have bases and heights correct. And then you do the integral over whatever interval it is. So with that let's try another one. In this example the base is an equilateral triangle and the cross sections are squares. So we make a diagram where one is the height, the original height. And if it's an equilateral triangle then the length on our x-axis is a square root of three. Which makes that hypotenuse equal to two. We have to develop now the equation for the length of the y-values. We have to develop an equation for the line between zero and the square root of three. And so that becomes y is equal to negative one over the square root of three. Remember we need our slope and we need our y-intercept. So negative one over the square root of three x plus one becomes the equation of that line. Now the area is what we need next. And remember the areas are going to be squares and the squares have a length of two y on each edge. So area is equal to two y quantity squared or four y squared. So now our volume remember is adding up all these little areas and creating a volume from that. So the volume is equal to four times the integral from zero to the square root of three of y squared. But now we'll put this in x's. So we'll have negative one over the square root of three x plus one quantity squared dx. If we put that into our calculator, I don't think we want to do this one by hand. If we put it into our calculator, we will get 2.309 cubic units. Let's change things a little bit. This time let's make the base sine x. And we're going to do our cross sections from zero to pi. And our cross sections are going to be equilateral triangles. So a little sketch of this, sine x from zero to pi looks like that. And of course the cross sections will be triangles this time. So you see in this case they do not go below the axis. They just stay above the axis. This length is y. Our area now has to be from an equilateral triangle whose base is a y. So setting this up, if this is y or this side is one half y, we have a 30, 60, 90 degree triangle again. So this is one half y. Then this side becomes the square root of three over two y. And that will be the height of the triangle. So the area of our triangle is one half y. And the height is square root of three over two y. So we get square root of three over four y squared. Now we have y is equal to sine x. So again we substitute in and we get square root of three over four sine squared x. Put that into our volume formula, the integral of the square root of three over four sine squared x. And don't forget your little delta x has to be in there. And we are going from zero to pi. Evaluating that you could either do it by hand or throw it on your calculator and you get an approximate value of 0.680 cubic units. And remember if you do the antiderivative of this, we have to separate it into the formula of half angle formulas and then do the antiderivative. So if you do it by hand, it's a little bit more complicated than just a simple sine cubed over three. Just remember that. Well let's go on to another problem. This time we have the base is a quarter of a circle with a radius of one and the cross section are squares. We love those squares don't we? So let's just take our base, quarter of a circle. The radius is one. So the formula for our circle is x squared plus y squared is equal to one. And if all the bases are squares, again I drew it perpendicular to the x-axis because that is usually the simplest way to do these. We will find that our square has a length of y. So our area is equal to y squared and we can transform that to one minus x squared. So once we do that we can do the volume is equal to the integral of one minus x squared dx. And since we only took the top of our circle as we were doing this, we can just go from zero to one on our interval. And in doing that this one's easily done by hand is x minus x cubed over three from zero to one. And that is equal to one minus one third which is equal to two thirds cubic units. So that one's not so bad. Let's try one more. This example says that the base is a circle with a radius of two and the cross sections are isosceles right triangles. So here we have a circle with a radius of two and now we need isosceles right triangles built on this base. And remember this the right angle is at the tip there. So if we took our isosceles right triangle and again it is built across the x-axis. So this is y and this is y. So the height on this would also be y. This is a 45, 45, 90 degree triangle. So if we want the area of this it's one half base which is two y times the height which is y so we get y squared. And if we transform that to x's if the base of the circle has a radius of two then the formula for our circle is x squared plus y squared is equal to four. So we have y squared is equal to four minus x squared. And in this one when we build our volume it's equal to four minus x squared. And we are going from negative to two. Or we can redo this and say the volume is equal to two times the integral from zero to two of four minus x squared dx. And once we integrate that we get two times four x minus x cubed over three. And we're going from zero to two. And we get a final answer of 32 over three cubic units. This is the last example so let me give you a few tips now on what to do. A rule of thumb. And I've done all my volumes like I said perpendicular to the x-axis. So when you do it perpendicular to the x-axis your area equation or area formula has to be an x's. So the volume is equal to the integral from a to b of a of x dx. If you do it perpendicular to the y-axis then your volume is equal to the integral from c to d where the integral values are of a of y dy. And some basics of doing cross sections. Draw the figure. Check which axis the cross sections are perpendicular to. Create the area using the base formulas. And then make sure you know that the volume is nothing more than the sum of all these areas. And of course that is an integral. So conclude your lesson on volumes using cross sections.