Question

Find the centroid of the region determined by the graphs of the inequalities. $$ y \leq 3 / \sqrt{x^{2}+9}, y \geq 0, x \geq-4, x \leq 4 $$

Short answer

The centroid (ave(x), ave(y)) can be found by evaluating the integrals in steps 2, 3 and 4.

Step-by-step solution

Set Up the Region

The region R can be illustrated on an x-y graph as the area bounded by the curve y = 3 / √(x² + 9), the x-axis (y = 0) and the vertical lines x = -4, x = 4.

Calculate the Area

To find the area (A) of the region R, we can use the formula \(A = \int_{a}^{b} f(x)dx\). Here, \(f(x) = 3 / \sqrt{x^{2}+9}\) and the integration limit a = -4 and b = 4. So, the area is \(A = \int_{-4}^{4} 3 / \sqrt{x^{2}+9} dx\).

Calculate the x-coordinate

The formula for ave(x) is \(1/A \int_{a}^{b} xdA\). Substituting our limits a (-4) and b(4), and our function for dA (3/\sqrt{x^{2}+9}), we get: \(ave(x) = \frac{1}{A} \int_{-4}^{4} x * (3 / \sqrt{x^{2}+9}) dx\).

Calculate the y-coordinate

The formula for ave(y) is \(1/2A \int_{a}^{b} (f(x))^2 dx\). Substitute our limits a (-4) and b(4), and our function for dA (3/\sqrt{x^{2}+9}), we get: \(ave(y) = \frac{1}{2A} \int_{-4}^{4} (3 / \sqrt{x^{2}+9})^2 dx\).

Evaluate the Integral

After setting up the integrals in step 2, 3 and 4, evaluate the integrals to get the area (A) and the x and y coordinates.