Week 11 and 12 - Applications of Integration

Computing Volumes

• Given a solid shape, if the areas of all slices orthogonal to a direction can be computed, then integrating that area function along the axis gives the volume.
• Volume of a solid of rotation can be computed by rotating a line around an axis to create shapes like cylinders or cones.

Solids of Revolution: Disk Method

• Given a region D in the plane between two functions $f(x)$ and $g(x)$, the goal is to compute the volume of the solid obtained by rotating D around the y-axis.

• Example: A circle centered at $(4, 3)$ with radius 1.

• Equation of the circle: $(x - 4)^2 + (y - 3)^2 = 1$

• Express $y$ in terms of $x$ to find two functions representing the upper and lower halves of the circle.
  $y - 3 = \pm \sqrt{1 - (x - 4)^2}$
  $y = 3 \pm \sqrt{1 - (x - 4)^2}$

• Thus, $f(x) = 3 + \sqrt{1 - (x - 4)^2}$ (upper half) and $g(x) = 3 - \sqrt{1 - (x - 4)^2}$ (lower half).

Volume Calculation by Cylindrical Shells

• Approximate the area by rectangles.
• Riemann sums: $\sum [f(x<em>i^<em>) - g(x</em>i^</em>)] \Delta x$
• Consider rotating a single rectangle around the y-axis, forming a cylindrical shell.
• Approximate this shell as a rectangular prism.

Dimensions of the Prism

• Width (circumference): $2 \pi x_i^*$
• Height: $f(x<em>i^<em>) - g(x</em>i^</em>)$
• Depth: $\Delta x$
• Volume of the cylindrical shell: $2 \pi x<em>i^* [f(x</em>i^<em>) - g(x_i^</em>)] \Delta x$

Riemann Sum and the Integral for Volume

• Summing the volumes of all cylindrical shells approximates the total volume of the rotational solid.
• This sum is a Riemann sum, leading to the integral:
  $V = \int_a^b 2 \pi x [f(x) - g(x)] dx$

Example Calculation

• Given $f(x) = \sqrt{x - 1} + 0.5$ and $g(x) = x - 1$ (corrected to be $g(x) = (x - 1)^2 + 0.5$ [Typo in transcript, should read -0.5])
• $\int (\sqrt{x - 1} - (x - 1)^2) 2 \pi x dx$
• Using the substitution $u = x - 1$, the integral can be solved.
• Resulting volume: $\frac{29}{30} \pi$

General Formula

• Region D bounded by $x = a$ and $x = b$, with $a$ and $b$ positive.
• Volume of the solid E obtained by rotating D around the y-axis:
  $V = \int_a^b 2 \pi x [f(x) - g(x)] dx$

Thin Rods and Center of Gravity

• Mass of a thin rod with density function $\rho(x)$ is computed by integrating the density function: $m = \int_a^b \rho(x) dx$
• If density is homogeneous (constant), then mass = density * length.

Center of Gravity Calculation

• For evenly distributed mass, the center of mass is at the midpoint.
• For non-uniformly distributed mass, the center of gravity is calculated as a weighted average.
• Consider the rod as a series of point masses and calculate the balancing point.

Derivation of Center of Gravity Formula

• With two masses $m<em>a$ at coordinate $a$ and $m</em>b$ at coordinate $b$, the center of gravity $x<em>B$ satisfies: $m</em>a (x<em>B - a) = m</em>b (b - x_B)$
• Therefore,
  $x<em>B = \frac{m</em>a a + m<em>b b}{m</em>a + m_b}$
• Generalizing for $n$ masses:
  $x<em>B = \frac{\sum m</em>i^* x<em>i^<em>}{\sum m</em>i^</em>}$

Continuous Mass Distribution

• Approximate the mass $m<em>i^<em>$ in each interval as $\rho(x</em>i^</em>) \Delta x$
• $x<em>B \approx \frac{\sum \rho(x</em>i^<em>) x<em>i^</em> \Delta x}{\sum \rho(x</em>i^*) \Delta x}$
• Taking the limit as $\Delta x \to 0$ yields:
  $x<em>B = \frac{\int</em>a^b x \rho(x) dx}{\int<em>a^b \rho(x) dx} = \frac{\int</em>a^b x \rho(x) dx}{M}$

Example: Homogeneous Rod

• If $\rho(x) = \rho<em>0$ (constant), then: $x</em>B = \frac{\int<em>a^b x \rho</em>0 dx}{\int<em>a^b \rho</em>0 dx} = \frac{\rho<em>0 \int</em>a^b x dx}{\rho<em>0 \int</em>a^b dx} = \frac{\frac{1}{2}(b^2 - a^2)}{b - a} = \frac{a + b}{2}$

Example: Non-Homogeneous Rod

• Rod between coordinates 2 and 3, with $\rho(x) = x^2 - 1$
• Mass: $M = \int_2^3 (x^2 - 1) dx = \frac{16}{3}$
• xB = \frac{\int2^3 x (x^2 - 1) dx}{M} = \frac{\int_2^3 (x^3 - x) dx}{\frac{16}{3}} = \frac{\frac{55}{4}}{\frac{16}{3}} = \frac{165}{64} > 2.5