Triple Integrals and Coordinates for Solids
Integrate over a solid region, and choose cylindrical or spherical coordinates when the solid has an axis or a centre of symmetry.
What a learner can do afterwards
- Set up and evaluate triple integrals whose limits describe a solid, including volume and constant-density mass
- Choose between cylindrical and spherical coordinates for a given solid
- State the volume element in each coordinate system and where its factors come from
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Integrating 1 over a solid gives its volume. For 0 <= x <= 3, 0 <= y <= 2 and 0 <= z <= 4, integrate dz first to get 4, then dy to get 8, then dx to get 24. Each differential belongs to its matching limits.
For a rectangular box, each side length is upper coordinate minus lower coordinate. Its volume is the product of those lengths. This also follows by integrating 1 successively over the three coordinate ranges, including ranges that do not start at zero.
Cylindrical coordinates use x = r cos(theta), y = r sin(theta), with z unchanged. The local lengths are dr, r dtheta and dz, so dV = r dz dr dtheta. The factor r comes from the polar area in the xy-plane.
Use rho >= 0, theta around the z-axis and phi measured from the positive z-axis, in radians. Then x = rho sin(phi) cos(theta), y = rho sin(phi) sin(theta), z = rho cos(phi). A full angular sweep uses 0 <= theta <= 2pi and 0 <= phi <= pi. Local lengths d rho, rho d phi and rho sin(phi) d theta give dV = rho² sin(phi) d rho d phi d theta.
A ball centred at the spherical-coordinate origin contains 0 <= rho <= R; rho = R is only its boundary. For radius 2, use rho from 0 to 2, phi from 0 to pi and theta from 0 to 2pi. Integrating rho² gives 8/3. Since an antiderivative of sin(phi) is -cos(phi), its integral is -cos(pi) + cos(0) = 1 + 1 = 2. The theta integral gives 2pi, so volume is (8/3) × 2 × 2pi = 32pi/3.
For a circular cylindrical tank centred on the z-axis with flat ends, cylindrical coordinates give constant radius, angle and height ranges. Spherical coordinates can describe it too, but usually need less convenient coupled bounds. Match the coordinates to the stated axis and boundaries.
When z is integrated first, project the solid onto the xy-plane for the outer two integrals. The inner z-limits run from the lower surface to the upper surface. For another inner variable, use the corresponding projection and span instead.
If the volumes do not match, check the integrand 1, coordinate conversion, Jacobian, limits and arithmetic. Agreement is useful but not proof that the region is right: different solids can have equal volume. Check the boundary inequalities independently too.
A constant density multiplies each tiny volume. Density 3 over 0 <= x <= 2, 0 <= y <= 1, 0 <= z <= 1 gives inner z-integral 3, middle y-integral 3 and outer x-integral 6. That result is mass, not volume.
A variable upper surface gives functional inner limits. For 0 <= x <= 2, 0 <= y <= 3 and 0 <= z <= x, integrating 1 in z gives x. Integrating in y gives 3x, then in x gives [3x²/2] from 0 to 2 = 6.
Convert the function as well as the volume element. In spherical coordinates, x² + y² + z² = rho². Multiply the converted function by rho² sin(phi) before integrating. The function and the volume factor have different roles.
Cylindrical coordinates can use another stated axis. Around the x-axis, the radial distance is √(y² + z²), and x remains the axial coordinate. Thus circular boundaries y² + z² = R² fit an x-axis cylindrical system.
Describe the solid layer by layer, match its symmetry to the coordinates, and check against known volume.
Lesson and assessment explanations written for Marble.
Additional questions and readings written for Marble. Videos credit Khan Academy.
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