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Volume with cross sections: triangle

This time, the cross section of our solid is given as the area between two curves.

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  • blobby green style avatar for user 1028646
    When am I ever going to need to use this
    (19 votes)
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  • male robot donald style avatar for user harry park
    Can someone explain to me what a Cross Section exactly is, and why do we need them?
    (5 votes)
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    • marcimus pink style avatar for user xyzPoKeFaNxyz
      ok it is very late, but i think it would help other people
      if you had a 3-D shape (your phone would be ok) and imagine that you cut it straight and what you find is a cross-section (ok, you may find this confusing, get an apple, cut it, and the juicy, yellowy side you find is the cross section!)
      To the answer to your question: as before, we have known how to find areas between curves, we are multiplying the infinitely small change of x (which is dx) by the area between the curves (which is f(x)-g(x)). First we need to find the function for the cross section which is true for any x, then we multiply the cross section (the area) times the width (which is infinitely small or dx) for every x, so we'll get the volume.
      (30 votes)
  • blobby green style avatar for user Gwyneth Dator
    I still don't understand why the area is (rad 2/2 h) x (rad 2/2 h) x (1/2)
    (10 votes)
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    • marcimus pink style avatar for user xyzPoKeFaNxyz
      Firstly, we use the Pythagorean theorem to find out both the length and height are h/rad(2) and if you multiply both numerator and denominator by rad(2), it is (rad 2/2 h).
      Secondly, we use the formula for area of triangles, which is height x width x 1/2, and plugging in what you have got, we have (rad 2/2 h) x (rad 2/2 h) x (1/2)
      (7 votes)
  • old spice man blue style avatar for user Mohammed Hamdi Elshafey
    How I can determine the upper curve and the lower curve?
    (3 votes)
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    • piceratops ultimate style avatar for user Darmon
      Looking at a graph is of course the best way. Here's a trick: for volume, it really does not matter which curve function you subtract from the other! All it will do is change the sign of your answer, which is irrelevant because volume is always positive (so if your answer has a negative sign, just make it positive). :)
      (17 votes)
  • leafers seed style avatar for user Hassaanghazali
    How do we know that the 3D shape will be a triangle? Could it also not be a curve? or a semicircle that is getting smaller and smaller?
    (3 votes)
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  • leafers seed style avatar for user Vlado Stoiloff
    Can we solve the area in this way?

    We are given the following sides of the isosceles triangle, as Sal has depicted:
    h = hypotenuse
    a1 = sqrt(2)/2*h
    a2 = sqrt(2)/2*h

    However we divide this triangle into two in order to find the height, which connects a1 and a2 and is perpinducular to the the hypotenuse.

    By doing so we actually having one of the legs as a new hypothenuse and half of the old hypotenuse and the old height as new adjacent and opposite side.

    Now we can use the Pytagorian Formula again on order to find the old height. That would be:

    (1/2*h)^2 + height^2 = (sqrt(2)/2*h)^2

    height^2 = (sqrt(2)/2*h)^2 - (1/2*h)^2
    height^2 = 2/4*h^2 - 1/4*h^2
    height^2 = 1/4 * h^2
    height = 1/2*h

    Area of new triangle:
    = 1/2*h * sqrt(2)/2*h * 1/2
    = sqrt(2)/8 * h^2

    Area of old triangle:
    = New triangle * 2
    = sqrt(2)/8 * h^2 *2
    = sqrt(2)/4 * h^2

    Which is NOT the same as Sal's result. I wonder what I did wrong?

    Thanks in advance!
    (4 votes)
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    • duskpin ultimate style avatar for user rahulmaru3507
      you were correct til the point where you found height = h/2. However, in the formula for the area of your half traingle, the sqrt(2)/2 side is NOT the base anymore, as it doesn't make a right angle with the new perpendicular line. what you can instead do is consider the whole original triangle, with base h, and height h/2 to get area = h^2/4
      (1 vote)
  • blobby green style avatar for user Tom Zhong
    Let B be the solid whose base is the circle x2+y2=9 and whose vertical cross sections perpendicular to the x-axis are equilateral triangles. Compute the volume of B. Can any one help me on this question?
    (1 vote)
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    • leafers seed style avatar for user Travis Bartholome
      I won't give you the answer, but I'll offer a general strategy for questions of that variety:

      1. Figure out which axis (and thus which variable) you'll be using for integration.
      2. Find an expression in terms of that variable for the width of the base at a given point along the axis.
      3. Find an expression for the area of the cross-section in terms of the base and/or the variable of integration.
      4. Integrate along the axis using the relevant bounds.

      A couple of hints for this particular problem:

      1. You know the cross-section is perpendicular to the x-axis. A width dx, then, should given you a cross-section with volume, and you can integrate dx and still be able to compute the area for the cross-section. (In essence: integrate dx.)

      2. Remember that to express a circle in terms of a single variable, you need two functions (one for above the x-axis and one for below it, in this case).

      3. The cross-section is an equilateral triangle, and you probably learned how to calculate the area for one of those long ago. You should have the base length from the previous step, which is all you need to find the cross-sectional area.

      4. Your bounds should obviously be the least and greatest x-values that lie on the circle.

      Hope all of that helps. Good luck.
      (6 votes)
  • duskpin ultimate style avatar for user Low2Key
    How will I know if the triangle is isosceles right triangle? Do they tell you in the question? Or do I have to figure it out on my own?
    (2 votes)
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  • blobby green style avatar for user weint101
    At , since the area = base * height *.5, wouldn't it be A= sqrt2/2 (h) * (h) *.5?
    (2 votes)
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    • hopper cool style avatar for user Iron Programming
      In this case the base and height were the same thing (we're dealing with an isosceles triangle);
      Base = (sqrt(2)/2) * h
      Height = (sqrt(2)/2) * h
      Area = (sqrt(2)/2) * h * (sqrt(2)/2) * h * (1/2)
      Simplify =>
      1. ((sqrt(2)*sqrt(2)/(2 * 2 * 2)) * (h*h)
      2. (2/8) * h^2
      3. (h^2/4)

      Hope this helps,
      - Convenient Colleague
      (2 votes)
  • blobby green style avatar for user benjaminpianoman
    Wouldn't there be an easier way to find the area of each "slice?"
    Instead of using 45-45-90 triangles, you could instead realize that since the triangle is half of the square, the height of the triangle is going to be 1/2h. Then multiplying 1/2h*h*1/2 to get the area of the triangle, you get the result 1/4h^2, which is the same as the result he got using the 45-45-90 triangles method
    (2 votes)
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Video transcript

- [Voiceover] Lets see if we can imagine a three-dimensional shape whose base could be viewed as this shaded in region between the graphs of Y is equal to F of X and Y is equal G of X. That's the base of our ... this purple, this I guess mauve or purple color is the base of it. It's kind of popping out of our screen. What I've drawn here in blue, you could view this kind of the top ridge of the figure. If you were to take cross sections of the figure, that's what this yellow line is. If you were to take cross sections of this figure, that our vertical, I should say our perpendicular to the X axis, those cross sections are going to be isosceles right triangles. This cross section is going to look like this, if you were going to flatten it out. So over here it's sitting. It's popping out of your page or out of your screen. If you were to actually flatten it out, the cross sections would look like this. It's going to be an isosceles right triangle with a hypotenuse of the isosceles right triangle sits along the base. Sits along the base. So, it's isosceles. So that's equal to that. It's a right triangle, and then this distance this distance between that point and this point is the same as the distance between F of X and G of X. F of X and G of X for this X value right over there. Now obviously that changes as we change our X value. To help us visualize this shape here, I've kind of drawn a picture of our coordinate plane. If we've viewed as an angle, if we're kind of above it, you can kind of start to see how this figure would look. Once again, I've drawn the base. I've drawn the base of it. I've drawn the base of it right over there. Maybe I should to make it clear. Let me get like this. Shade it in kind of parallel to these cross sections. So, I've drawn the base right over there. There's two other sides. There the side that's on ... I've drawn it here. I guess you could view it on its top side or the left side right over there. Over on this picture that would be this. When we're looking at it from above. Then you have this other side. I guess on this view this one you could call kind of the right side. Over here this is kind of the ... when you viewed over here this is the bottom side. The whole reason why I set this up, and we tempting to visualize this figure. I want to see if you can come up with a definite integral that describes the volume of this figure. That kind of almost looks like a football if you cut it in half or a rugby ball. It's skewed a little bit as well. What's an expression a definite integral that expresses the volume of this. I encourage you to use the fact that it intersects at the points. These functions intersect at the point zero zero and (c,d). Can you come up with some expression a definite in row of terms of zeros, and Cs, and Ds, and Fs, and Gs that describe the volume of this figure? Assuming you've paused the video, and have had a go at it, let's think about it. If we want to find the volume, one way to think about it is we could take the volume of, we could approximate the volume as the volume of these individual triangles. That would be the area of each of these triangles times some very small depth. Some very small depth. I'll just shade it in to show the depth. Some very small depth which we could call DX. Once again we could find the volume of each of these by finding the area, the cross sectional area there, and then multiplying that times a little DX. A little DX which would give us three, so this is a little DX, which would give us three-dimensional. That's our DX. I could write that a little bit neater. DX to give us a little bit of three-dimensional depth. How could we ... what is the volume of one of these figures going to look like? Well, if we say, let me just call this height H. We know that H is going to be F of X minus G of X. That's this distance right over here. Let's call that H. We know that H is going to be, maybe I should say H of X. It is going to be a function of X. H of X is going to be F of X. F of X minus G of X. Minus G of X. Given a H, what is going to be the area of this triangle? This is a 45, 45, 90 triangle. This is 90, then this is going to have to be 45 degrees. That's going to have to be 45 degrees. We know that the sides of a 45, 45, 90 triangle or squared of two times the hypotenuse. This is going to be squared of two over two times the hypotenuse. Squared two over two time the hypotenuse. You could get that straight from the pythagorean theorem. If the side, let's say it's length A, then this side has length A. You're going to have A squared plus A squared is equal to the hypotenuse squared, or two A squared is equal to the hypotenuse squared. A squared is equal to the hypotenuse squared over two, or that A is equal to H over the squared of two, which is the same thing as a squared of two H over two. I just rationalized the denominator, multiplied by open numerator, and then denominator by squared of two. That's where I got this from. What is the area going to be? The area is just going to be your base times your height times one-half. Let me write that down. The area the area there the area is just going to be the base, which is squared of two over two times our hypotenuse, times the height which is squared of two over two time our hypotenuse, times one-half. Times one-half. If we didn't do that one-half, we'd be figuring out the area of this entire square. This is obviously where were concerned with the triangle. What's this going to be? This is going to be squared of two over two times squared of two over two is going to be one-half and then you're going to have another one-half. So, it's one-fourth, one-fourth H square. Did I do that right? This is ... yes it's going to be two over four, which is one-half, and then times another one-half. This one-fourth H squared is the area. Now, what's going to be the volume of each of these triangles? Well the volume of each of these triangles right over here, the volume is just going to be our area times DX. It's just going to be one-fourth H squared times our depth. Times our depth DX. If we just integrated a bunch of these, from our X equals zero all the way to X equals C, we essentially have our volume of the entire figure. So how could we write that? We want to write. We want to essentially to find the volume of the figure. We're kind of in the home stretch here. Actually let me write volume. This is volume of a section. Volume of a cross section. So, what's the volume of the entire thing going to be? The volume volume of the figure of the figure is going to be the definite integral, the definite integral from X equals zero to X equals C. X equals zero to C of one-fourth H squared. One-fourth. We know that H is equal to F of X minus G of X. Instead of H, I'm going to write F of X minus G of X squared squared DX. DX and we're done. We just found an expression for, a definite integral expression I guess you could say, for the volume of this strange figure that we have defined.