 Hi everyone. These are a few examples for you on how to find volumes of solids by using cross sections. So the first scenario we're going to look at asks us to find the volume of the solid whose base is bounded by the graphs of y equals x plus one and y equals x squared minus one with the indicated cross sections taken perpendicular to the x-axis. So the fact that we're having cross sections perpendicular to the x-axis, right away you know this is going to be dx integrals that we set up. So the first one we want to look at are when the cross sections are squares. So let's go ahead and graph the functions just so we can get a sense of what the base is going to look like. So if you have the equations under y equals it should be sufficient to do zoom six to graph it. So there's the linear function x plus one and here comes the quadratic x squared minus one. So that region in between the two is the base of the solid and you need to envision the solid itself rising up towards you from that base. Now the cross sections we're going to take perpendicular to the x-axis in this first example are squares. So let's go back and let's draw ourselves a sketch of what this looks like. So there's the linear function our cross sections perpendicular to the x-axis. So think once again of your representative rectangle and that rectangle could go anywhere as long as it's touching both curves but it's because of that orientation that vertical orientation of your rectangle that it's a dx problem. So the first thing we want to do is find the length of that representative rectangle the length of our base and to do that we can do top minus bottom. So on the top it's hitting the curve x plus one on the bottom it's hitting x squared minus one so if we simplify that we'd have x minus x squared plus two. So that is the length of the side of the square that is going to be our cross section. So the next thing we want to do is come up with an expression for the area of the square. Well if the base is x minus x squared plus two obviously the base squared will be the area of the square. So then to get our volume of this solid that's created we need to do the integral of that area expression and we do need to find these points of intersection. You can do those graphically and find those graphically on your calculator. Remember that they are x values so go ahead and try them you should get negative one and a positive two. So then if you evaluate that on your calculator or by hand by finding the antiderivative you should find that your answer is 8.1. So get that a try to make sure you get the right answer. So the second example we're going to look at uses the same equations the same type of solid but now we want our cross sections to be rectangles of height one. Now they're still perpendicular to the x axis so that length of the base really does remain the same but now we need the area of the rectangle. So obviously area of a rectangle is the base times the height well our expression for the length of the base we just found and our height is just simply one in this case. So to get our volume then our limits of integration are still the same. We're going to take the integral of our area of the type of cross section we had and once again you can either evaluate that by hand by finding your antiderivative or you can evaluate on your calculator. So give it a shot and you should hopefully come out with 4.5 as your answer. Now as we saw with areas there really are no units of measure given in this case if there were they would be cubic units because they are volume answers that we're finding. So if you did need to attach units it would just be a generic cubic units that you would attach. So the third one we're going to look at once again is still the same base but now we want our cross sections to be equilateral triangles. So let's go back to our picture for a second. So imagine that representative rectangle being the base of the equilateral triangle and the triangle is rising out towards you up towards you and what you're looking at here in the picture is the base of that equilateral triangle. So if you envision that in your head the solid then would have like a pointy top to it. So the length of the base still remains the same. Now our area of an equilateral triangle, remember the formula for finding the area of an equilateral triangle is square root of 3 over 4 times the base squared. So this is the base so we have root 3 over 4 times that quantity squared. So that's what we're going to take the volume of the integral of. Of course that root 3 over 4 you could put to the very front and just evaluate the integral itself and then multiply that answer by root 3 over 4. However you care to do it is up to you. Give it a shot. The answer you should get is 3.507. Now an interesting thing to note, notice that part of this integral is really what we had back here when we did the volume by using square cross sections. So this is kind of interesting if you think about it. What you're finding is that there is a relationship between the volume of the solid found with cross sections that are equilateral triangles versus the volume of the solid found with cross sections of squares. So basically you could take this answer we had from the cross sections being squares, multiply that by root 3 over 4 and you will get this answer over here. So kind of an interesting relationship between the two. So let's now look at another scenario with different types of equations as our base. And this time we want our cross sections to be perpendicular to the y-axis. So remember that means we're going to have dy expressions. So let's go back to the calculator and graph this. So you can zoom 6 to graph it. It's just a simple cubic equation. You might want to zoom in a little bit. So the solid and its base are formed by this curve, the line y equals 0. So that's the x-axis and then a vertical line at x equals 1. So the base had almost a right triangle shape to it where the hypotenuse is concave in a little bit. So let's go ahead and draw ourselves a picture of what this looks like. So here would be your line at x equals 1. So since our representative rectangles are now going perpendicular to the y-axis, they are horizontal therefore. That's what's making it a dy problem. So we otherwise follow the same process. The first thing we want to do is find the length of that representative rectangle that is our base. This time though, since it's going horizontal, we want to go right minus left as opposed to top minus bottom. So on the right side, it's hitting the vertical line 1. On the left side, it's hitting the curve. Now remember though, it's a dy problem. We cannot use x's. So we need to take this equation and rewrite it x in terms of y. And that's how we want to work with it. So that's the length of that representative rectangle from left to right. So the area then of our square, so imagine that rectangle being the base of your square and the square rising up towards you. So the area is simply going to be that base, the quantity squared. Our volume then is going to be the integral of the area Now our limits of integration now have to be y values. So the lower limit is zero, upper limit is one. So you can either evaluate that by hand by finding the anti-derivative or go ahead and do it on your calculator. You should get that it's point one. It rounds to point one, I know. So we have two more to look at using this same setup. So the next one we're going to look at, we now want our cross sections perpendicular to the y-axis to be isosceles triangles of height four. For isosceles triangles, you're simply going to use that typical one-half base times height. And you're given the height, the base is what we already found, the one minus y to the one-third. So the length of the base remains the same. The area of our isosceles triangle, we're simply going to use one-half base times height. So one-half the base times the height of four. And of course we can simplify that. Our volume then is going to be the integral from zero to one of that area expression. The two of course could move to the front if you wish. And go ahead, evaluate that, and you should get rounding off to point five. Finally the last one we're going to look at are our cross sections being semicircles. So the length of the base is still the same. Now let's go back to the picture for a second. So that rectangle you see that is the base of your semicircles. So you have a semicircle rising from that base up towards you. The length of that though is really the diameter of the semicircle. So you have to think of how you get area of a circle, a semicircle. Well area of a circle is pi r squared. We need to take half of that to get the area of the semicircle. But you have to remember that one minus y to the one-third is the diameter. You need to cut that in half to get the radius. So as we set this up, using one half pi r squared, we have one half pi. The radius we're going to get by doing one minus y to the third divided in two, that quantity squared. Now you can simplify this if you wish. You don't necessarily have to. You're squaring this denominator of two to make four times the one-half in the front, you get pi over eight. So that simplifies a little bit easier for when you go to type it in and evaluate your integral. Your volume then, you can put the pi over eight out front if you wish, and then your integral from zero to one. Which you'll notice, again this integral here is the one we had set up to find the volume of the solid when our cross sections were squares. So if you notice that, you could just take that previous answer and multiply it by pi over eight. So if you want to give that a shot, you should get in the end 0.039 approximately.