Question

Find the volume of the solid generated by revolving the region bounded by the graphs of the equations about the y-axis.
y=3(2−x)
y=0
x=0

Answer 1

To find the volume I suppose calculus integration should be involved but first we solve for x in the equation

y=3(2-x)

y=6-3x

x=$\frac{6-y}{3}$

Then introducing the boundaries which are when y=0 this mean the volume will not be valid therefore using boundaries when x=0 then y=6.

using the formula V=π $\int\limits^1_0 {(3(2-x))^{2} } \, dx$

then

V= $\pi \int\limits^6_0 [{\frac{(6-y)}{3} ] ^{2} } \, dx$

V= $\frac{\pi }{9}$[36y-$\frac{y^{3} }{3}$]    bounded by o and 6

V =$\frac{\pi }{9} (36(6)-\frac{(6)^{3} }{3})-0$

v= 50.265 $units^{3}$

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