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## Integral Calculus

### Unit 3: Lesson 8

Volume: triangles and semicircles cross sections

# Volume with cross sections: triangle

AP.CALC:
CHA‑5 (EU)
,
CHA‑5.B (LO)
,
CHA‑5.B.2 (EK)
This time, the cross section of our solid is given as the area between two curves.

## Want to join the conversation?

• When am I ever going to need to use this
(1 vote)
• Maybe the real treasure was the friends you made along the way
• Can someone explain to me what a Cross Section exactly is, and why do we need them?
• 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.
• I still don't understand why the area is (rad 2/2 h) x (rad 2/2 h) x (1/2)
• 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)
• How I can determine the upper curve and the lower curve?
• 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). :)
• 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?
• That the cross-section is a triangle is part of the given information in the problem. That was not deduced from the shape of the base because, as you said, a shape with the given base could have any kind of cross-sections.
• 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)
• 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.
• At , since the area = base * height *.5, wouldn't it be A= sqrt2/2 (h) * (h) *.5?
• 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
• Why is the height equal to the base here, isn't it smaller?
• In this case Sal chose ℎ to represent the base. The height is ℎ∕2.
(1 vote)
• Why can't he just use 1/2(f(x)-g(x))^2? How come he does the thing with rad(2)/(2)?
• 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?