Question

In Exercises \(31-40,\) find \(M_{x}, M_{y},\) and \((\bar{x}, \bar{y})\) for the laminas of uniform density \(\rho\) bounded by the graphs of the equations. $$ y=\sqrt{x}, y=0, x=4 $$

Short answer

The first moments about the x- and y-axes of the lamina are \(M_{x}=8\rho\) and \(M_{y}=8\rho\) respectively, and the centroid of the lamina is located at \((\bar{x}, \bar{y}) = \left(\frac{3}{2}\rho, \frac{3}{2}\rho\right)\)

Step-by-step solution

Determine the region

Firstly, the region R for the lamina must be determined. This region is bounded by the graphs of the three equations: \(y=\sqrt{x}\), \(y=0\), and \(x=4\). The region R is thus the area under the curve \(y=\sqrt{x}\) from \(x=0\) to \(x=4\).

Setup the integrals for \(M_x\) and \(M_y\)

Next, the double integrals for the first moments \(M_x\) and \(M_y\) are set up. As the density \(\rho\) is uniform, it can be considered as a constant factor outside the integral. The first moment about the x-axis \(M_x\) is given by the double integral of y over the region R: \[M_x = \rho\int_0^4\int_0^{\sqrt{x}} y \,dy\,dx\] and the first moment about the y-axis \(M_y\) is given by the double integral of x over the region R: \[M_y = \rho\int_0^4\int_0^{\sqrt{x}} x \,dy\,dx\]

Evaluate the integrals

The double integrals for \(M_x\) and \(M_y\) are then evaluated. This gives the first moments about the x- and y-axes respectively. Using the properties of integrals, these can be simplified as follows: \[\begin{align*} M_x & = \rho\int_0^4 \frac{1}{2}x \,dx = \frac{\rho}{2} \int_0^4x \,dx = \frac{\rho}{2} \cdot \frac{x^2}{2}\Bigg|_0^4 = \frac{16\rho}{2} = 8\rho \ M_y & = \rho\int_0^4 x \,dx = \rho \cdot \frac{x^2}{2}\Bigg|_0^4 = \frac{16\rho}{2} = 8\rho \end{align*}\]

Calculate the Area

The total area A of the region R is calculated from the integral of 1 over the same region: \[A = \int_0^4\int_0^{\sqrt{x}} \,dy\,dx = \int_0^4 \sqrt{x} \,dx = \frac{2}{3}x^{3/2}\Bigg|_0^4 = \frac{16}{3}\]

Calculate Centroid Coordinates

Finally, the coordinates of the centroid \((\bar{x}, \bar{y})\) are calculated by dividing the first moments by the total area: \[\bar{x} = \frac{M_y}{A} = \frac{8\rho}{16/3} = \frac{3}{2}\rho, \quad \bar{y} = \frac{M_x}{A} = \frac{8\rho}{16/3} = \frac{3}{2}\rho\] Note that the density \(\rho\) cancels out because it is uniform.