 Welcome back everyone to the final part of lecture five about volumes using cross-sectional slicing I want to look at one more example again these things can be kind of challenging So it's good to have lots of examples to see how one can approach this thing So let's describe the solid in question and then try to find the volume of said solid So imagine a wedge is cut out of a circular cylinder of radius four by two planes Okay, and then there's an illustration of what's going on here down below which is courtesy of James Stewart's calculus textbook one of the planes is just going to be perpendicular to the axis of the cylinder so basically like you see the illustration right here if you think of the bottom circle of this cylinder I Should use a different color as white blends too much with the background if you think of this bottom right here as the XY plane That's one of the planes. We're going to use the slices thing And then the other plane is going to come through the x-axis of the XY plane at a 30 degree angle or our pie Sixth if you prefer radians and such as it like this and so let me kind of erase what I wrote here And so you try to see that there's this wedge that's cut out of The cylinder so we're trying to describe this region right here Okay, and so it's going to come about by using cross sectional slices Which if we look at cross sections, and so let's give it a little bit of description of the base So from a different perspective, we're looking at the x-axis right here The y-axis comes up right here as well like so and Then we're going to have a semicircle That lives inside this XY plane like so and so what we're trying to do is we're trying to describe this Cymicircle right here now as the circle the base circles can have radius of four That was given up in the description right here. This circle will have the equation y equals the square root of 16 minus x squared Like so so this point right here would be four comma zero This point right here would be negative four comma zero in terms of this description All right, so this kind of gives us the base of our of our solid and what we're going to be doing is stacking on top of this base Cross sections perpendicular to the x-axis. So we're seeing cross sections being stacked On the base like this and the height of these cross sections will depend on where we are in the semicircle But the cross sections are going to look like right triangles Right triangles resembling this shape over here and we've chosen these right triangles so that the angle right here It's going to be a 30 degree angle All right, and so these these cross sections Let me clean this thing up a little bit If we now focus on the cross section right here, right we have to find the area of The area of a triangle. It's a right triangle So the area is just going to be one half base times height where here is your base and here is your height The base is just going to be the y-coordinate So we're going to get one half y Times the height. How can we figure out the height of this thing? Well, that's where the 30 degree angle comes into play A tangent ratio would be very useful here tangent of 30 degrees is going to equal Opposite over adjacent the opposite side is h the adjacent side is y Clearing the denominators. We actually see that the height will equal y times tangent of 30 degrees And if you don't remember that off the top of your head Or you don't have a calculator. Hopefully you remember sine and cosine of 30 degrees, right? So tangent is sine of 30 degrees over cosine of 30 degrees Sine of 30 degrees is one half And cosine of 30 degrees is root three over two. So if you simplify that thing you end up with y over the square root of three As so and and so that's going to be plugging in For the the height of our of our triangle over here So we get y over the square root of three Now some people have a pathological fear of having square roots in the denominator So if you do need to rationalize it, I suppose You can do so if you really feel necessary So if you were to rationalize the denominator, you would end up with like a root three Over this over six. I don't see any need to have to do such a thing right there So we're just going to leave it the way it is So basically what I'm saying is the area is going to equal y squared over two times the square root of three That's going to be the area of one of these one of these Triangles, but again, this is in terms of y if we set up our integral The volume of this wedge is going to be the integral of an area function a of x dx And let's try to identify what the bounds of integration would be if we come back up here Our x values is allowed to range anywhere in this spectrum right here from negative four to four So those are going to be the bounds of integration We end up with x is negative four to four We'll notice that's again a symmetric interval. It feels very likely that we're going to be able to Use symmetry to simplify the integral here. We'll get back to that in just a moment The area function which we saw from before y squared over two squared of three we can plug that in right here So we end up with two times the integral from zero to four We use symmetry y squared over two times the square root of three dx So we can simplify things a little bit like the two out in front cancels with this two right here But we have to still deal with the y squared. How do we deal with y squared? We'll come up to the semicircle we had before Because the this distance y Given by this point x comma y This y is constrained by this function right here y equals 16 the square root of 16 minus x squared therefore y square will be 16 minus x squared So if we insert that into The integral here We end up with the integral with one over the square root of three Sorry one over the square root of three times integral from zero to four We get 16 minus x squared dx And so integrate that thing The anti-derivative Will look like well we get the constant one over square three in front We're going to get 16 x minus x cubed over three As we go from zero to four Plugging in the four we're going to get 16 times four which is 64 minus 64 over three If you fact out the 64 which is common to everything 64 over the square of three We're left with one minus one minus one-third The difference which is two-thirds And so then in the end we would get 64 times two Which is 128 Over the square root of three times three so three root three And so this would give us the exact volume of this wedge that we slice out of That cylinder from above we could approximate this if we wanted to but I will I'll we'll be happy with exact values with we know how to throw this number into a calculator and go from there And so that brings us to the end of lecture five. I appreciate everyone's uh watching through these videos here about Finding volumes using cross sectional analysis using these integrals here If you like these videos, please do click the like button Feel free to subscribe so you learn more about videos that are created in the future You have any questions about any of the content either in this lecture in previous videos Or in any part of this series feel free to post your comment your questions in the comments here on youtube And I'll be glad to reply to them and answer any questions you might have Check out the next video and learn some more about volume And how we can use integrals to calculate volume of solids. I'll see you then. Bye everyone