The idea here is that we want to take very thin "slices" perpendicular to the axis we are rotating around. These slices will function as radii that we plug into the area function of the circle, A = Tr². We then can get volumes by multiplying by little changes (dx or dy). Finally, we sum over these volumes (which are very thin!) to get the volume of the entire cone. So, we need to slice from y = 0 to y = x, meaning the radius at each point is given by 0 to x = h to get the integral x – 0 = x. We then integrate from x = ch So Area(circular slices) dæ which would be T(x)² dx. I will leave it to you to evaluate this integral to get the desired result.
The idea here is that we want to take very thin "slices" perpendicular to the axis we are rotating around. These slices will function as radii that we plug into the area function of the circle, A = Tr². We then can get volumes by multiplying by little changes (dx or dy). Finally, we sum over these volumes (which are very thin!) to get the volume of the entire cone. So, we need to slice from y = 0 to y = x, meaning the radius at each point is given by 0 to x = h to get the integral x – 0 = x. We then integrate from x = ch So Area(circular slices) dæ which would be T(x)² dx. I will leave it to you to evaluate this integral to get the desired result.
Elementary Geometry For College Students, 7e
7th Edition
ISBN:9781337614085
Author:Alexander, Daniel C.; Koeberlein, Geralyn M.
Publisher:Alexander, Daniel C.; Koeberlein, Geralyn M.
Chapter10: Analytic Geometry
Section10.1: The Rectangular Coordinate System
Problem 40E: Find the exact volume of the solid that results when the region bounded in quadrant I by the axes...
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
Transcribed Image Text:We are going to walk through two ways to derive the volume of a right circular cone.
To do so, we start by considering the line y = ;x.
We then consider the region bounded by this line and the lines y = 0 and x = h - note that
this is a right triangle with base h and height r.
The idea is that if we rotate this region about the x axis, we will get a filled-in right circular cone
with radius r and height h.

Transcribed Image Text:The idea here is that we want to take very thin "slices" perpendicular to the axis we are rotating
around. These slices will function as radii that we plug into the area function of the circle, A =
Tr. We then can get volumes by multiplying by little changes (dx or dy). Finally, we sum over
these volumes (which are very thin!) to get the volume of the entire cone.
So, we need to slice from y = 0 to y = x, meaning the radius at each point is given by
= 0 to x
x. We then integrate from x
h to get the integral
So Area(circular slices) dæ which would be T(x)² dx. I will leave it to you to evaluate
this integral to get the desired result.
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