Answer:
V = (1/3)πr²h
Step-by-step explanation:
The volume of a cone is 1/3 the volume of a cylinder with the same radius and height.
Cylinder Volume = πr²h
Cone Volume = (1/3)πr²h
where r is the radius (of the base), and h is the height perpendicular to the circular base.
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<em>Comment on area and volume in general</em>
You will note the presence of the factor πr² in these formulas. This is the area of the circular base of the object. That is, the volume is the product of the area of the base and the height. In general terms, ...
V = Bh . . . . . for an object with congruent parallel "bases"
V = (1/3)Bh . . . . . for a pointed object with base area B.
This is the case for any cylinder or prism, even if the parallel bases are not aligned with each other. (That is, it works for oblique prisms, too.)
Note that the cone, a pointed version of a cylinder, has 1/3 the volume. This is true also of any pointed objects in which the horizontal dimensions are proportional to the vertical dimensions*. (That is, this formula (1/3Bh), works for any right- or oblique pyramid-like object.)
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* in this discussion, we have assumed the base is in a horizontal plane, and the height is measured vertically from that plane. Of course, any orientation is possible.
-54-7r=10+25r
-7r-25r=10+54
-32r=64
r=64/-32
r=-2
In order to do this, you must first find the "cross product" of these vectors. To do that, we can use several methods. To simplify this first, I suggest you compute:
‹1, -1, 1› × ‹0, 1, 1›
You are interested in vectors orthogonal to the originals, which don't change when you scale them. Using 0,-1,1 is much easier than 6s and 7s.
So what methods are there to compute this? You can review them here (or presumably in your class notes or textbook):
http://en.wikipedia.org/wiki/Cross_produ...
In addition to these methods, sometimes I like to set up:
‹1, -1, 1› • ‹a, b, c› = 0
‹0, 1, 1› • ‹a, b, c› = 0
That is the dot product, and having these dot products equal zero guarantees orthogonality. You can convert that to:
a - b + c = 0
b + c = 0
This is two equations, three unknowns, so you can solve it with one free parameter:
b = -c
a = c - b = -2c
The computation, regardless of method, yields:
‹1, -1, 1› × ‹0, 1, 1› = ‹-2, -1, 1›
The above method, solving equations, works because you'd just plug in c=1 to obtain this solution. However, it is not a unit vector. There will always be two unit vectors (if you find one, then its negative will be the other of course). To find the unit vector, we need to find the magnitude of our vector:
|| ‹-2, -1, 1› || = √( (-2)² + (-1)² + (1)² ) = √( 4 + 1 + 1 ) = √6
Then we divide that vector by its magnitude to yield one solution:
‹ -2/√6 , -1/√6 , 1/√6 ›
And take the negative for the other:
‹ 2/√6 , 1/√6 , -1/√6 ›
