AP®︎ Calculus AB (2017 edition)
The volume of a solid with semi-circular cross sections and a triangular base.
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- I still don't get how 1-x is diameter. Isn't it suppose to be radius?(74 votes)
- I was confused for a while too because I thought this triangle was rotated around the x-axis to create a half cone. Don't understand why he used this shape as an example(8 votes)
- In the beginning of the video, Khan explained that the cross section that is perpendicular to the x axis will form a semi circle, but I am still having trouble seeing that a semi circle will be the 3D cross section for this figure. Why can't it be a triangle or a square?(25 votes)
- I think the semi circle is just an arbitrary shape that he has chosen for this example. You could make the cross sections triangular or square if you wanted to and still use the same base shape. That would just be a different example of the same general principle.(16 votes)
- Yeah I still don't get this😤(8 votes)
- is the volume of a cone not 1/3 pirh? When you solve shouldn't you get 3 in the denom not 4?(2 votes)
- I thought the same thing as you at first. But after reading some comments I realized this is NOT a cone shape. In a perfect cone the cross sections are circular only when the cross section is taken PERPENDICULAR to the center line of the cone. In this video the semi circular cross sections are not perpendicular to the center line but perpendicular to the lower edge of the shape (represented by the x axis). This is only clear once you pay attention to the yellow lines in the left and right drawings. Then it becomes clear that 1) the y value is the diameter, 2) the x axis is the lower edge and 3) the yellow line is the upper edge of the base that lies flat/horizontal. The center line of the shape on the right would run from the points (0,.5) to (1,0).(14 votes)
- You say that y = 1 - x = Dia. or 2r ? If you view three dimensional view I see that 1 - x = y = radius and not dia. or 2 r. Everything else is well explained and understood. Please help me to understand that y = Dia. and not radius. If I would be able to see that y = dia. then radius is (1 - x)/2.(7 votes)
- 1-x is the base the semicircle (the diameter), not it's height (the radius).follow the yellow line it will be clear to you then.(5 votes)
- I've seen a couple of people explain that "1-x" is the "base (diameter)" of the semicircle, and therefore NOT the radius. I apologize for a redundant question, but I don't understand why. Visually, it seems very clear to me that "1-x" is half the diameter of a full circle "D/2", which would make "1-x" the radius. Doing "(1-x)/2" makes it seem like you're actually taking half of the radius. Could you help clear this up?(3 votes)
- I looked at the video and see why you were confused. Initially I did visualize the "slice" from the x axis to the line
y = 1 - xas being pivoted from the x axis out into the pos and neg z direction to fill out 3-space. But I looked at the tent-looking diagram of rotation effect again and realized this critter lives inside of one octant - positive x's positive y's and positive z's. Look again at the tent diagram and notice where the x and y axes are and it might pop at you rightly too. we were both thinking of this as a rotational solid. In a sense it is. The rotational axis would be a median of the right triangle - the one that goes from the vertex on the x axis to the midpoint of the side on the y axis.(5 votes)
- Right at the beginning at1:20, when Sal encouraged us to figure out the volume of the shape, I used a different method that still gave me the correct answer. Can anyone please tell me if the method I used (shown below) is wrong?
I first recognized the shape to be half a cone. The volume to calculate the volume of a cone is (pi*r^2*h)/3. To get half a cone, I divided that by 2, giving me the formula ((pi*r^2*h)/3)/2. From here on, I only need to substitute in the values for r and h to get the correct answer.
To get the height, I saw that in the diagram of the cone, you can see that the height of the cone is the same as the one of the bases of the triangle formed by the function f(x) in the first quadrant. Since the formula for f(x) is x+y=1, I figured the base had to be 1.
To calculate the radius, I noticed that another base of the triangle formed by the function f(x) in the first quadrant was the diameter of the cone. Again, since the formula for f(x) is x+y=1, I figured the base diameter had to be 1. Radius = diameter by 2. Radius = 1/2.
Substituting numbers back into original equation:
=((1/4)pi /3) /2
=(1/12 pi) /2
= 1/24 pi
Can someone please tell me if this method is wrong and if so, what is my error?(2 votes)
- It is wrong because you made assumption for the shape, and the height. Just because a diagram APPEARS to be a certain way, does not mean it is, you need to follow the information given (formula, labelled heights, lengths, etc).
So yes, it was wrong, because it will not always work unless the given information fits your assumptions, which may very well happen, but why chance it?(6 votes)
- I am having a hard time visualizing this. Is the semi-circle rising out of the page towards me in the third dimension? Is it shaped like a funnel, with the diameter being tiniest as we approach x =1?(3 votes)
- Hi! I had the same problem before too. The question will actually tell you what kind the shape the base is and that's how you can imagine. Your problem is why the f(x) will suddenly become the diameter of a semicircle, right? It is actually the question that gives us the answer(4 votes)
- what are examples of solid?(2 votes)
- Geometrical regular solids are pyramids, prisms, and other polyhedrons, cylinders, cones, torus, and spheres. Other (irregular) three-dimensional figures can be solids, if their functions are bounded in each of the three dimensions.(3 votes)
- [Voiceover] This right over here is the graph of x plus y is equal to one. Let's say the region that's below this graph but still in the first quadrant, that this is the base of a three dimensional figure. So this region right over here is the base of a three dimensional figure. What we know about this three dimensional figure is if we were to take cross sections that are perpendicular to the x axis, so cross sections I guess we could say that are parallel to the y axis, so let's say we were taking a cross-section like that, we know that this will be a semi-circle. If we take a slightly different view of the same three-dimensional, we would see something like this. I've laid the coordinate plane down flat and I'm looking at it from above. This cross-section, if we're looking at it at an angle, and if the figure were transparent, it would be this cross-section right over here. It would be that cross-section right over here, which is a semi-circle. If we were to take this cross section right over here, along the y axis, that would be this cross section. This larger semi-circle. Given what I've just told you, I encourage you to pause the video and see if you can figure out the volume of this thing that I've shaded in a little bit. The volume of this thing that I've just started. This three dimensional figure that we are attempting to visualize. I'm assuming you've had a go at it. One way to think about it is, well let's just think about each of these, let's split up the figure into a bunch of discs. If you figure out the volume of each of those discs, and sum them up then that would be a pretty good approximation for the volume of the whole thing. Then, if you took the limit as you get an infinite number of discs that are infinitely thin, then you're going to get the exact volume. Let's just take the approximating case first. Let's say that right over here at x, what is going to be the diameter of the disc at x. Well, to think about that we just have to re-express x plus y is equal to one. That's the same thing as saying that the function that F of x or y is equal to F of x is equal to one minus x. The diameter of this circle right over here, let me make it clear, the diameter of this circle is going to be this height, is going to be the difference between one minus x and the x axis, or between one minus x and x equals zero. This is just going to be, as a function of x, the diameter's going to be one minus x. Now, if you want to find the surface area for a circle, or if we want to find the area of a circle, we know that area is pi r-squared. For our semi-circle, you're going to divide by two. So what's the radius going to be? Just let me zoom in a little bit. What is the radius of one of these semi-circles going to be? The radius I could draw it like this. My best attempt to draw one of these things. It's gonna look something like this. I'm trying to draw it at an angle. The radius, what's gonna be half the diameter? The diameter is one minus x. The radius is going to be one minus x over two. Right, that distance is one minus x over two. That distance is one minus x over two. That distance is one minus x over two. So, that is one minus x over two is equal to the radius. What would be the area of of this side right over here? Well, if it was a full circle, it would be pi r-squared, but it's a semi-circle so it's pi r-squared over two. Let me write this, so the area is equal to pi r-squared over two 'cause it's a semi-circle. In terms of x, our area as a function of x right over there, let me write it that way, our area as a function of x, is gonna be pi over two. It's gonna be pi over two times one minus x over two squared. It's one minus x over two squared. If I wanted a volume of just this disc right over here, I, then, multiply that area times the depth. I just multiply the area times this depth here. We could call that d x or we could call that delta x. We could call the depth there delta x. Actually, let me be clear, that's the area. The volume of one of those shells is going to be equal to, and I'll just write this in one color, is gonna be pi over two times one minus x over two squared, times the depth. The area times the depth. This is volume of one of these half discs. The volume of the whole thing we can approximate is the sum of these, or we can take the limit is our delta x's get much, much, much, much, smaller and we have an infinite number of these things, which is essentially, if we're taking that limit, we're gonna take the definite integral. We could take the definite integral, so the volume that we care about, the volume of this figure, is going to be equal to the definite integral from x equals zero to x equals one. That's where we intersect the x axis. X equals zero to x equals one of, we're integrating an infinite number of these things that are infinitely thin. It's going to be pi over two times, what's one minus x-squared? Let me just expand it out for fun right over here. That's the same thing as x minus one squared. So it's going to be x squared minus two x plus one, and then two squared is four, over four. Instead of delta x, I'm going to right d x now. I'm gonna write d x because I'm taking the limit as these become infinitely small and I have an infinite nine. I'm summing an infinite number of them. The volume is just going to be, we just have to evaluate this definite integral. If you feel so inspired, feel free to pause and try to evaluate this definite integral. Let's just take some of these constants out. Let's take pi over eight out. Our volume is going to be equal to pi over eight times the definite integral from zero to one of x-squared minus two x plus one d x, which is equal to pi over eight. The antiderivative of this is x to the third over three minus x-squared plus x. We're going to evaluate that at zero and one. This is going to be equal to pi over eight. When you evaluate it at one, you get one third minus one plus one. It's going to be one third. When you evaluate at zero, you get zero minus zero plus zero. It's going to be minus zero so it's just pi over eight times one third which is equal to pi over 24 and we are done.