Volume of Cone z = √(x²+y²) under z = 4

Calculus 3 · Axiom Academy

EXAMPLE Volume of the Cone under z=4 Finding the volume trapped between a cone and a flat cap with a triple integral in cylindrical coordinates. Find the volume of the solid region bounded below by the cone and above by the plane z=4 . Cross-section in the r–z plane. The cone becomes the line z=r ; it meets the cap z=4 where r=4 . For a fixed r , the solid runs vertically from z=r up to z=4 — this is the height we integrate. Nicely done. You set up and evaluated a triple integral for the volume between a cone and a flat cap — a textbook case of cylindrical symmetry. Convert to cylindrical: since x^2+y^2=r^2 , the cone simplifies to the clean line z=r . Read the bounds off the geometry: for a fixed r , z runs from the cone up to the cap, ; the cone meets the cap at r=4 , and sweeps 0 to . Never drop the Jacobian: in cylindrical coordinates the volume element is — the extra r is what makes the integral correct. Result: cubic units, which matches the cone-volume formula . The same recipe — convert, read the vertical bounds off a cross-section, remember the r — handles almost any solid with an axis of symmetry.

This is the written version of the interactive lesson above. See the full Calculus 3 course.