[attachment=0:bfh13tp7]789.zip[/attachment:bfh13tp7]
A abihoward New member Joined Oct 21, 2010 Messages 7 Nov 16, 2010 #1 [attachment=0:bfh13tp7]789.zip[/attachment:bfh13tp7]
G galactus Super Moderator Staff member Joined Sep 28, 2005 Messages 7,216 Nov 16, 2010 #2 Find the volume of the donut-shaped region (torus) given by ρ=4sin2(ϕ)\displaystyle {\rho}=4sin^{2}(\phi)ρ=4sin2(ϕ) centered at the origin. One way is to use a triple integral.
Find the volume of the donut-shaped region (torus) given by ρ=4sin2(ϕ)\displaystyle {\rho}=4sin^{2}(\phi)ρ=4sin2(ϕ) centered at the origin. One way is to use a triple integral.
G galactus Super Moderator Staff member Joined Sep 28, 2005 Messages 7,216 Nov 17, 2010 #3 Try a triple integral in spherical coordinates: ∫02π∫0π∫04sin2ϕρ2sinϕ dρdϕdθ\displaystyle \int_{0}^{2\pi}\int_{0}^{\pi}\int_{0}^{4sin^{2}{\phi}}{\rho}^{2}sin{\phi} \;\ d{\rho}d{\phi}d{\theta}∫02π∫0π∫04sin2ϕρ2sinϕ dρdϕdθ
Try a triple integral in spherical coordinates: ∫02π∫0π∫04sin2ϕρ2sinϕ dρdϕdθ\displaystyle \int_{0}^{2\pi}\int_{0}^{\pi}\int_{0}^{4sin^{2}{\phi}}{\rho}^{2}sin{\phi} \;\ d{\rho}d{\phi}d{\theta}∫02π∫0π∫04sin2ϕρ2sinϕ dρdϕdθ
