How do I find the volume of the solid generated by revolving the region bounded by #y=x^2#, #y=0#, and #x=2# about the #x#-axis? The #y#-axis?

1 Answer
May 19, 2018

1)#Volume=piint_0^2x^4*dx=(32/5)pi (unite)^3#

2)#Volume=piint_0^4[(2^2)_2-(sqrty^2)_1]*dy=piint_0^4[4-y]*dy=8pi (unite)^3#

Explanation:

the rose region is revolving about the x-axis and y-axis

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1)when the shaded region revolving a bout x-axis

#Volume=piint_a^by^2*dx#

#Volume=piint_0^2y^2*dx=piint_0^2x^4*dx=pi[1/5*x^5]_0^2#

#=pi[(32/5)-0]=(32/5)pi (unite)^3#

2)when the shaded region revolving about the y-axis

#Volume=piint_d^c[(x^2)_2-(x^2)_1]*dy#

#Volume=piint_0^4[(2^2)_2-(sqrty^2)_1]*dy#

#=piint_0^4[4-y]*dy=pi[4y-1/2y^2]_0^4#

#=pi[16-8]=8pi (unite)^3#