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?

Question

Samantha Adkison · Accepted Answer

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#

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

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#

Aurora Alvarado · Answer

undefined To 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 and the y-axis, you can use the method of cylindrical shells for the x-axis and the method of disks or washers for the y-axis.

1. Around the x-axis:
   Using cylindrical shells, the volume $ V $ is given by:
   $$ V = \int_{0}^{2} 2\pi x \cdot x^2 \, dx $$
   Evaluate this integral to find the volume.

2. Around the y-axis:
   Using disks or washers, the volume $ V $ is given by:
   $$ V = \pi \int_{0}^{4} x^2 \, dy $$
   Here, $ x^2 $ represents the radius of the disk or washer, and $ dy $ represents an infinitesimally small change in the height.
   First, express $ x $ in terms of $ y $ and then evaluate the integral to find the volume.