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Answer 1 · 2017-12-25T19:07:24

#9pi#

while you can use shell method, i think it's easier to rewrite y=x as a function of y and use disc method to rotate about the y-axis.

as a function of y, y=x becomes x=y.

using disc method, the volume of the solid is: #piint_a^b(r(y))^2dy#

a and b are the upper and lower bounds of integration. in this problem, #b=3# and #a=0# (because x=y intersects the axis of rotation at #(0,0)#)

r(y) is the distance between the function and the axis of rotation. in this problem, #r(y)=y-0=y#

plugging in: volume=#piint_0^3y^2dy#
#=pi(F(3)-F(0))#, where #F(y)=1/3y^3# or the antiderivative of #y^2#.

#=pi(1/3(3)^3-1/3(0)^3)#
#=pi(9-0)=9pi#