The problem involves finding the center of mass of a homogeneous lamina, which lies between $x = \frac{\pi}{2}$ and $x = \frac{3\pi}{4}$, with the curve $y = \sin(x)$ as the upper boundary and the $x$-axis ($y = 0$) as the lower boundary.

The center of mass (or centroid) $( \bar{x}, \bar{y} )$ of a region bounded by curves can be found using the following formulas:

• The $x$-coordinate of the centroid, $\bar{x}$, is given by:

$\bar{x} = \frac{1}{A} \int_{a}^{b} x f(x) \, dx$

• The $y$-coordinate of the centroid, $\bar{y}$, is given by:

$\bar{y} = \frac{1}{2A} \int_{a}^{b} [f(x)]^2 \, dx$

• The area $A$ of the region is:

$A = \int_{a}^{b} f(x) \, dx$

Here, $f(x) = \sin(x)$, and the region is bounded between $x = \frac{\pi}{2}$ and $x = \frac{3\pi}{4}$. Let’s calculate these one by one.

Step 1: Calculate the Area $A$

$A = \int_{\frac{\pi}{2}}^{\frac{3\pi}{4}} \sin(x) \, dx$

Let me compute the area first.The area $A$ of the lamina is $\frac{\sqrt{2}}{2}$.

Step 2: Calculate $\bar{x}$ (the $x$-coordinate of the center of mass)

$\bar{x} = \frac{1}{A} \int_{\frac{\pi}{2}}^{\frac{3\pi}{4}} x \sin(x) \, dx$

I'll now compute $\bar{x}$.The $x$-coordinate of the center of mass is:

$\bar{x} = -\sqrt{2} + 1 + \frac{3\pi}{4}$

Step 3: Calculate $\bar{y}$ (the $y$-coordinate of the center of mass)

$\bar{y} = \frac{1}{2A} \int_{\frac{\pi}{2}}^{\frac{3\pi}{4}} \sin^2(x) \, dx$

I'll compute $\bar{y}$ next.The $y$-coordinate of the center of mass is:

$\bar{y} = \frac{\sqrt{2}(2 + \pi)}{16}$

Final Answer:

The center of mass (centroid) of the lamina is located at: $\left( \bar{x}, \bar{y} \right) = \left( -\sqrt{2} + 1 + \frac{3\pi}{4}, \frac{\sqrt{2}(2 + \pi)}{16} \right)$

Would you like more details or have any questions about this solution?

Here are five related questions to explore:

1. How would the center of mass change if the region were bounded by a different function, such as $\cos(x)$?
2. What is the physical interpretation of the center of mass in this context?
3. How would you compute the moment of inertia for this lamina?
4. What happens to the center of mass if we extend the limits beyond $\frac{3\pi}{4}$?
5. How does the homogeneity of the lamina affect the calculation of the center of mass?

Tip: The integral of trigonometric functions can often be simplified by using trigonometric identities or substitution techniques!