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Jasty
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Please I need a respectable proof how to get the volume of the truncated cone. I need it really quick. So please could you help me. No numbers just "the method" how to get that formula. Thanks.
Jasty said:Please I need a respectable proof how to get the volume of the truncated cone. I need it really quick. So please could you help me. No numbers just "the method" how to get that formula. Thanks.
HallsofIvy said:Now imagine the entire cone, divided into thin disks: each has thickness "dz" and radius, x= (R1-R2)z/H+ R2 so area [itex]\pi [(R1-R2)z/H+ R2)^2[/itex] and volume [itex]\pi [(R1-R2)z/H+ R2)^2dx[/itex]. To find the entire volume integrate that from z= 0 to z= H.
The formula for finding the volume of a truncated cone is V = (1/3)πh (r2 + rR + R2), where h is the height of the cone, r is the radius of the smaller base, and R is the radius of the larger base.
The formula is derived by using the method of integration to calculate the volume of the conical frustum, which is the shape created when the top of a cone is cut off parallel to the base. This leads to the formula V = (1/3)πh (r2 + rR + R2).
A cone is a three-dimensional shape with a circular base and a curved surface that tapers to a point at the top. A truncated cone, also known as a frustum, is a cone with its top removed parallel to the base, creating a flat top surface. It is essentially a cone with two circular bases of different sizes.
The formula can only be applied to truncated cones, but the concept of calculating the volume of a shape by using the method of integration can be applied to other shapes as well.
Knowing the volume of a truncated cone can be useful in various real-world applications, such as calculating the volume of a storage tank or the amount of material needed to fill a conical mold for a product. It is also an important concept in mathematics and science, demonstrating the use of integration to find the volume of irregular shapes.