Question

84.a. Find the center of mass of a thin plate of constant density covering the region between the curve  y=1/√x  and the x-axis from  x=1 to x=16.

Accepted Answer

Identify the region and the density: The plate has constant density, so we can denote the density as $$ \rho  (a $$constant). The region is bounded by the curve $$ y = \frac{1}{\sqrt{x}} $$, the x-axis $$ y=0 $$, and the vertical lines $$ x=1 $$ and $$ x=16 . $$Set up the expressions for the area and moments: Since the density is constant, the mass $$ M  is $$proportional to the area of the region. The area $$ A  is $$given by the integral $$ A = \int_{1}^{16} \frac{1}{\sqrt{x}} \, dx . $$Find the coordinates of the center of mass $$ (\bar{x}, \bar{y}) $$: Use the formulas for the centroid of a lamina with constant density:

$$ \bar{x} = \frac{1}{A} \int_{1}^{16} x \cdot \frac{1}{\sqrt{x}} \, dx $$

$$ \bar{y} = \frac{1}{2A} \int_{1}^{16} \left( \frac{1}{\sqrt{x}} \right)^2 \, dx $$

Note that $$ \bar{y} $$ uses the average height of the region, which is the moment about the x-axis divided by the area. Evaluate the integrals step-by-step: 
- For the area $$ A $$, integrate $$ \int_{1}^{16} x^{-1/2} \, dx .
- $$For $$ \bar{x} $$, integrate $$ \int_{1}^{16} x^{1/2} \, dx .
- $$For $$ \bar{y} $$, integrate $$ \int_{1}^{16} x^{-1} \, dx . $$After computing the integrals, substitute the results back into the formulas for $$ \bar{x} $$ and $$ \bar{y}  to $$find the coordinates of the center of mass.

84.a. Find the center of mass of a thin plate of constant density covering the region between the curve y=1/√x and the x-axis from x=1 to x=16.

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Key Concepts

Center of Mass for a Lamina

Definite Integrals for Area and Moments

Function and Region Description