Measuring volume of spheres using triple integrals

robertjford80 · May 28, 2012

Homework Statement

Two 3D solids for volume exercises: left, a paraboloid cap under z = 4 - (x^2 + y^2) above z = (x^2 + y^2)^2 - 1; right, a cone-like solid with z = 1 - r and z = -sqrt(1 - r^2) with axes x, y, z labelled.

I'm just interested in knowing where the 4 comes from in front of the integral.

tiny-tim · May 28, 2012

hi robertjford80!

that 4∫₀^π/2 is really ∫₀^2π …

i suppose the writer thought it looks confusingly like ∫₀⁰, though i see nothing wrong with it! :redface:

robertjford80 · May 28, 2012

Ok, pretty unexpected but I guess it makes sense.

HallsofIvy · May 28, 2012

In other words, they are using the circular symmetry of the figures, integrating from 0 to [itex]\pi/2[/itex], rather than from 0 to [itex]2\pi[/itex], then multiplying by 4.

Measuring volume of spheres using triple integrals

Homework Statement

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