S synx New member Joined Feb 25, 2006 Messages 20 Nov 18, 2006 #1 using shells method: y = x^(1/2) bounded by x=4, y=0 revolved around y-axis V=2pi * integral(x*x^(1/2)dx) from 0 to 4 Is this right?
using shells method: y = x^(1/2) bounded by x=4, y=0 revolved around y-axis V=2pi * integral(x*x^(1/2)dx) from 0 to 4 Is this right?
G galactus Super Moderator Staff member Joined Sep 28, 2005 Messages 7,203 Nov 18, 2006 #2 Which axis are you revolving it about?.
G galactus Super Moderator Staff member Joined Sep 28, 2005 Messages 7,203 Nov 18, 2006 #4 Looks OK. \(\displaystyle \L\\2{\pi}\int_{0}^{4}x^{\frac{3}{2}}dx=\frac{128{\pi}}{5}\) Try it with washers, see if you get the same thing. \(\displaystyle \L\\{\pi}\int_{0}^{2}(16-y^{4})dy\)
Looks OK. \(\displaystyle \L\\2{\pi}\int_{0}^{4}x^{\frac{3}{2}}dx=\frac{128{\pi}}{5}\) Try it with washers, see if you get the same thing. \(\displaystyle \L\\{\pi}\int_{0}^{2}(16-y^{4})dy\)
