# SOLUTION: A dentist places a gold filling in the shape of a cylinder with a hemispherical top in a patient's tooth. The radius of the filling is 3 mm. The height of the cylindrical portion i

Algebra ->  Algebra  -> Bodies-in-space -> SOLUTION: A dentist places a gold filling in the shape of a cylinder with a hemispherical top in a patient's tooth. The radius of the filling is 3 mm. The height of the cylindrical portion i      Log On

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 Question 37171: A dentist places a gold filling in the shape of a cylinder with a hemispherical top in a patient's tooth. The radius of the filling is 3 mm. The height of the cylindrical portion is 4 mm. If dental gold costs \$16.00 per cubic millimeter, how much did the gold for the filling cost? Round to the nearest cent. Use π ≈ 3.14. Answer by fractalier(2101)   (Show Source): You can put this solution on YOUR website!We will calculate the volumes of the cylinder and hemisphere separately, then add them. V(cyl) = (pi)r^2h = (pi)(3^2)(4) = 36(pi) V(hemi) = .5(4/3)(pi)r^3 = .5(4/3)(pi)(3^3) = 18(pi) Total volume is 54(pi) = 169.56 cubic mm Now multiply by \$16 per cubic mm and the answer is 169.56 x 16 = \$2712.96 And you wanted a gold filling! Yikes!