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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: Yoda on May 19, 2014, 07:19:42 pm
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A hemispherical bowl can be described as the solid of revolution generated by rotating x^2+y^2=a^2 about the y-axis for-a<=y<=0. The bowl is filled with water.At time t=0, water starts running out of a small holein the bottom of the bowl, so that the depth of water in the bowl at time t is h cm. The rate at which the volume is decreasing is proportional to h.
Show that when the depth is h, the volume, Vcm^3, of water remaining is V=pi(a x h^2-1/3 x h^3) where 0<h<=a.