Let R be the region in the first quadrant bounded by the graph of #y=8-x^(3/2)#, the x-axis, and the y-axis. What is the best approximation of the volume of the solid generated when R is revolved about the x-axis?

1 Answer
Jan 20, 2018

The volume of the solid generated when R is revolved about the #x#-axis is #848/15pi# or approximately #177.6#.

Explanation:

The area of a circle with a radius equal to #8 - x^(3/2)# at any given #x# is #pi*(8 - x^(3/2))^2#.

#pi * (8 - x^(3/2))^2#

#pi * (64 - 16x^(3/2) + x^3)#

#pix^3 - pi16x^(3/2) + 64pi#

We wish to figure out exactly what region we're looking at. This can be done by looking at the #x#-intercept of #f(x)#.

#8 - x^(3/2) = 0#

#8 = x^(3/2)#

#8^(2/3)=x#

#4 = x#

It has an #x#-intercept of #4#.

With this, we know to integrate #pix^3 - pi16x^(3/2) + 64pi# from #0# to #4#.

The antiderivative of #pix^3 - pi16x^(3/2) + 64pi# is:

#pi/3 color(red)x^2 - (32pi)/5 color(red)x^(5/2) + 64pi color(red)x#

So,

#int_0^4 (pix^3 - pi16x^(3/2) + 64pi)dx#

#= pi/3 color(red)4^2 - (32pi)/5 color(red)4^(5/2) + 64pi color(red)4 - (pi/3 color(red)0^2 - (32pi)/5 color(red)0^(5/2) + 64pi color(red)0)#

#= pi/3 color(red)4^2 - (32pi)/5 color(red)4^(5/2) + 64pi color(red)4#

#= pi*(1/3 color(red)4^2 - (32)/5 color(red)4^(5/2) + 64* color(red)4)#

#= pi * (16 / 3 - 1024/5 + 256)#

#= pi * (80 / 15 - 3072 / 15 + 3840 / 15)#

#= 848/15pi#

#~~ 177.6#