Respuesta :
Answer:
V = 2*pi*R^2*H / 3 .... Accurate according to formula
Step-by-step explanation:
Given:
- Coordinates of the triangular region:
( 0 , 0 ) , ( 0 , R ) , ( H , R )
- The region is rotated about x-axis.
Find:
Use the Washer Method to find the volume of the solid obtained when R is rotated about the x axis.
Use the Shell Method to find the volume of the solid obtained when R is rotated about the x axis
Is there a way to check the accuracy of the answers to questions (1) and (2) using the formula for the volume of a cone?
Solution:
- Using washer method. Define functions f(x) and g(x), such that f(x) > g(x):
f(x) = R
g(x) = ( R / H )*x
- Set up the integral for washer method as follows:
[tex]V = \pi *\int\limits^H_0 {[f(x)^2 - g(x)^2]} \, dx\\\\V = \pi *\int\limits^H_0 {[R^2 - (R/H)^2*x^2]} \, dx\\\\V = \pi {[R^2*x - \frac{R^2}{3*H^2} *x^3]} \limits^H_0\\\\V = \pi {[R^2*H - \frac{R^2*H}{3}]} \\\\V = \frac{2*\pi*R^2*H }{3}[/tex]
- Using the shell method we have to express the area A as a function of y first:
A(y) = 2*pi*Radius*Width
A(y) = 2*pi*R*H - 2*pi*y*(H-H*y/R)
A(y) = 2*pi*R*H - 2*pi*(H*y-H*y^2/R)
- Set up the integral and evaluate:
[tex]V = \int\limits^a_b {A(y)} \, dy\\\\V = \int\limits^R_0 [{2*pi*y*H - 2*pi*(H*y -\frac{H*y^2}{R} )}] \, dy\\\\V = [{pi*H*y^2 - 2*pi*(H*y^2 / 2 -\frac{H*y^3}{R} ) }] \limits^R_0\\\\V = [{pi*R^2*H - 2*pi*(H*R^2 / 2 -\frac{H*R^3}{3R} )}] \\\\V = pi*R^2*H - \frac{1}{3}* \pi*R^2*H\\\\V = \frac{2*\pi * R^2 * H}{3}[/tex]
- The volume of the solid can be verified by removing a cone from a cylinder as follows:
V = V_cylinder - V_cone
V = pi*R^2*H - pi*R^2*H / 3
V = 2*pi*R^2*H / 3
- The volume of solid derived is accurate according to the formulas.
