Let R be the region enclosed by the graphs of #y=(64x)^(1/4)# and #y=x#. How do you find the volume of the solid generated when region R is revolved about the x-axis?

Question

Let R be the region enclosed by the graphs of #y=(64x)^(1/4)# and #y=x#.  How do you find the volume of the solid generated when region R is revolved about the x-axis?

Evelyn Agard · Accepted Answer

#(64pi)/3#

First, plot the two graphs, shading the bounded region between them.

Second, recall the formula for solid of revolution about the #x#-axis

#int_(x_0)^(x_f)pi("outer radius")^2-pi("inner radius")2dx#

The outer radius is the height from the #x#-axis to the outer most graph, or #y=(64x)^(1/4)#

#int_(x_0)^(x_f)pi((64x)^(1/4))^2-pi("inner radius")^2dx#

The inner radius is the height from the #x#-axis to the inner most graph, or #y=x#

#int_(x_0)^(x_f)pi((64x)^(1/4))^2-pi(x)^2dx#

The lower limit of integration is the smallest #x#-value where the two curves meet, or #x_0=0#. The upper limit of integration is the largest #x#-value where the two curves meet, or #x_f=4#

#int_(0)^(4)pi((64x)^(1/4))^2-pi(x)^2dx#

#int_(0)^(4)pi(64x)^(1/2)-pix^2dx#

#int_(0)^(4)pi8x^(1/2)-pix^2dx#

#int_(0)^(4)pi8x^(1/2)dx-int_(0)^(4)pix^2dx#

#8piint_(0)^(4)x^(1/2)dx-piint_(0)^(4)x^2dx#

#16/3pi[x^(3/2)]_(0)^(4)-pi/3[x^3]_(0)^(4)#

#16/3pi(4)^(3/2)-pi/3(4)^3=(64pi)/3#

Axel Alvarez · Answer

undefined To find the volume of the solid generated when region R is revolved about the x-axis, you can use the method of cylindrical shells. First, determine the limits of integration by finding the intersection points of the two curves $y = (64x)^{1/4}$ and $y = x$. Solve $ (64x)^{1/4} = x$ for $x$ to find the limits. Once you have the limits, set up the integral $\int_{a}^{b} 2\pi x(f(x) - g(x)) dx$, where $f(x)$ is the upper function and $g(x)$ is the lower function. Integrate from $a$ to $b$ to find the volume of the solid of revolution.