Center of Mass

Introduction to Coordinate Systems in Integration

  • Purpose: Exploring a different coordinate system for integrals.

Understanding the Setup

  • Begin with the equation of a circular cylinder:

    • x2+y2=4x^2 + y^2 = 4

    • The cylinder has a radius of 2.

    • Domain of x: 2x2-2 \leq x \leq 2

    • y values lead to both positive and negative square roots of y.

    • Area of integration is within the cylinder, represented in the xy-plane.

  • Note: The footprint (area) defined by x2+y2=4x^2 + y^2 = 4 is not part of the solid over which we are integrating.

Visual Representation

  • Visualizing the cylinder:

    • It encompasses all points inside a circle of radius 2 on the xy-plane.

  • Analysis of z values:

    • Rearranging gives:

    • z=2±4x2y2z = 2 \pm \sqrt{4 - x^2 - y^2}

    • The equation represents a sphere centered at another point. Specifically, it translates to:

    • Relation: x2+y2+(z2)2=4x^2 + y^2 + (z - 2)^2 = 4

    • This shape resembles a ball positioned above the origin, resting on the xy-plane.

Coordinate Transformation

  • Spherical Coordinates:

    • Transitioning into spherical coordinates involves using:

    • x=ρsinϕcosθx = \rho \sin \phi \cos \theta

    • y=ρsinϕsinθy = \rho \sin \phi \sin \theta

    • z=ρcosϕz = \rho \cos \phi

  • Upon squaring the coordinates, the simplification leads to:

    • ρ24z=0\rho^2 - 4z = 0

  • Transforming to spherical coordinates gives us the equation:

    • ρ=4cosϕ\rho = 4 \cos \phi

    • Which describes a sphere on the xy-plane.

Graphical Representation in Spherical Coordinates

  • When analyzing the angle phi:

    • At ϕ=0\phi = 0, ρ=4\rho = 4 which is at the top of the sphere.

    • As ϕ\phi increases, ρ\rho shrinks, demonstrating the volume of integration.

  • Limitations:

    • Avoid values where ρ\rho becomes negative; valid limits only exist if the sphere remains above the xy-plane.

Integration in Spherical Coordinates

  • Importance of maintaining the volume element for integration:

    • dV=ρ2sinϕdρdϕdθdV = \rho^2 \sin \phi \, d\rho \, d\phi \, d\theta

  • Function for integration, since x2+y2+z2x^2 + y^2 + z^2 translates into spherical coordinates as:

    • ρ2\rho^2

  • The integral involves calculating:

    • ρ3dV\int \rho^3 \, dV

    • Result: 409621π\frac{4096}{21} \pi (simplified result of integration).

Triple and Double Integrals in Applications

  • Starting point: Integrals can be applied to find mass when combined with density functions.

  • Density Function (Coefficient of mass density):

    • Notation: δ\delta for two-dimensional analysis.

    • Indicates mass density in area (grams per square centimeter).

  • The double integral of a density function gives the total mass:

    • For mass calculation, consider tiny rectangle areas, massdensity×areamass \approx density \times area.

Mass Moments and Center of Mass

  • First Mass Moment:

    • Notated as MyM_y, represents value weighted by distance from y-axis.

  • Moment statistical description:

    • Involves density and distance tallies.

    • Approach: Integrate the density function weighted by the respective coordinate (x or y).

  • To find the center of mass:

    • Use the relations xˉ=M<em>ym\bar{x} = \frac{M<em>y}{m} and yˉ=M</em>xm\bar{y} = \frac{M</em>x}{m} where m is the total mass.

Example Problem Demonstration

  • Consider the curve y=xy = x and the area between 0x10\leq x \leq 1.

  • Density function is proportional to distance from the y-axis (thus kxkx).

  • Integrals needed to find mass and moments:

    • m=k01(x2)dxm = k \int_0^1 (x^2)dx.

    • Final calculation indicates k12\frac{k}{12} for mass.

  • Calculation involves repeated integration for moments, leading to coordinates of the center of mass:

    • Result: Center of mass located at (35,12)\left(\frac{3}{5}, \frac{1}{2} \right)

Moving into Three-Dimensional Problems

  • Explore triple integrals focusing on volumes, specifically tetrahedrons:

    • Example defined by density function over a triangular base.

  • Integrations must be set up correctly to account for volume:

    • mass=δ(x,y,z)dzdydxmass = \int \int \int \delta(x,y,z) \, dz \, dy \, dx

    • For valid regions, arrange appropriate limits based on the geometric shape.

  • Employ density functions to approximate mass accurately - yields real-life applications in physics and engineering contexts.

Conclusion

  • Review forming integrals in various coordinate systems, recognizing changes in geometry.

  • Maintain clarity in definitions and transformations for smooth calculations in multiple dimensions.