How to Find the Volume Between a Hemisphere and Cone | Triple Integral Calculus 3 Tutorial
#calculus3 #integral #hemisphere #cone
In this step-by-step tutorial, we use multivariable calculus and spherical coordinates to calculate the volume between a hemisphere and a cone. The solid is confined within the sphere x2+y2+z2=36x^2 + y^2 + z^2 = 36x2+y2+z2=36, above the xy-plane, and outside the cone z=4x2+y2z = 4 \sqrt{x^2 + y^2}z=4x2+y2. This volume cannot be determined using basic geometry due to the curved surface of the cone.
By leveraging spherical coordinates (ρ,θ,ϕ\rho, \theta, \phiρ,θ,ϕ), we simplify the integration process and set the bounds for ρ\rhoρ (0 to 6), θ\thetaθ (0 to 2π2\pi2π), and ϕ\phiϕ (from arctan(1/4)\arctan(1/4)arctan(1/4) to π/2\pi/2π/2). After setting up and evaluating the triple integral, we arrive at the exact volume of the region.
This video demonstrates the power of spherical coordinates in solving complex volume problems. If you're studying Calculus 3 and triple integrals, this tutorial will guide you through the process with clarity and precision.
Chapters / Timestamps
00:00 Introduction
00:36 Description of solid region
01:17 3 Dimensional Plot of Solid Region
04:22 Cone surface in spherical coordinates
06:48 Tangent function trigonometry, cone at constant phi
10:00 Declination angle bounds
15:15 Multivariable Calculus conventions for z(x,y) surfaces
17:41 Theta bounds azimuthal circle
17:57 Rho radius bounds
20:20 Spherical volume element and integrand
21:11 Split up the triple integral
21:31 Trigonometric integral, bounds, trig inverse trig right triangle
23:45 Final answer, volume of solid
24:14 Summary
25:04 Take action!
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