Question 1208484
We use the Washer Method for all of these.
a)
Note that we only consider the part above the x-axis. (the region below the x-axis gives the same results). This means that {{{y=4\sqrt(x)}}}. This gives the volume as {{{pi*int((4\sqrt(x))^2,dx,0,4)=pi*int(16x,dx,0,4)=pi*(8*4^2-8*0^2)=128pi}}}.
b)
We integrate with respect to {{{y}}} this time, since we're rotating about the y-axis. Note that at {{{x=4}}}, {{{y=8}}}. (again, we're only considering the part above the x-axis) The outer radius is always {{{4}}}, and the inner radius for a given {{{y}}} is the corresponding x-value, which is {{{y^2/16}}}. This means that the volume is {{{pi*int(4^2-(y^2/16)^2,dy,0,8)=pi*((16*8-8^5/1280)-(16*0-0^5/1280))=512/5*pi}}}.
c)
We also integrate with respect to {{{y}}}. For a given {{{y}}}, the radius is {{{4-x}}} (no outer/inner radius this time), or {{{4-y^2/16}}}. This means the volume is {{{pi*int((4-y^2/16)^2,dy,0,8)=pi*int(y^4/256-y^2/2+16,dy,0,8)=pi*((8^5/1280-8^3/6+16*8)-(0^5/1280-0^3/6+16*0))=1024/15*pi}}}.
d)
We integrate with respect to {{{y}}}. The inner radius is {{{4}}}, and the outer radius for a given {{{y}}} is {{{8-x}}}, or {{{8-y^2/16}}}. This means the volume is {{{pi*int((8-y^2/16)^2-4^2,dy,0,8)=pi*int(y^4/256-y^2+48,dy,0,8)=pi*((8^5/1280-8^3/3+48*8)-(0^5/1280-0^3/3+48*0))=3584/15*pi}}}.