Question

Set up an integral for both orders of integration, and use the more convenient order to evaluate the integral over the region \(R\). \(\int_{R} \int x d A\) \(R\) : sector of a circle in the first quadrant bounded by \(y=\sqrt{25-x^{2}}, 3 x-4 y=0, y=0\)

Short answer

The more convenient integral to compute was the integral over y followed by x (i.e., \(\int_{0}^{3} \int_{0}^{\sqrt{25-x^2}} x \, dy \, dx\)), resulting in an integral value of \(\frac{18}{5}\).

Step-by-step solution

Understanding the Region R

The region R is defined by three curves: \(y=\sqrt{25-x^2}\) (the upper semicircle of radius 5), \(3x-4y=0\), a line which gives \(y=\frac{3}{4}x\) and \(y=0\) (the x-axis). Since we're told it's in the first quadrant, that means our x and y values are both positive.

Define the limits of integration

The x-boundaries of the sector are from 0 to where the line intersects with the semicircle. This point can be found by setting \(\sqrt{25-x^2} = \frac{3}{4}x\). Solving this equation gives \(x=3\). The y-boundaries also extend from 0 to the semi-circle, which is \(y=\sqrt{25-x^2}\) So the limits of integration are \(0 ≤ x ≤ 3\) and \(0 ≤ y ≤ \sqrt{25-x^2}\).

Setup the Integral in both Orders

1. Integral over y, then x: \(\int_{0}^{3} \int_{0}^{\sqrt{25-x^2}} x \, dy \, dx\). 2. Integral over x, then y: \(\int_{0}^{5} \int_{\frac{4}{3}y}^{ \sqrt{25-y^2}} x \, dx \, dy\). These integrals represent the same region R but with different orders of integration.

Select the More Convenient Integral

The integral \(\int_{0}^{3} \int_{0}^{\sqrt{25-x^2}} x \, dy \, dx\) appears to be easier to compute as the inner integral doesn't involve x and can be directly integrated to \(xy\) between the limits 0 and \(\sqrt{25-x^2}\).

Evaluate the Integral

We can now carry out the integration: \(\int_{0}^{3} x [\sqrt{25-x^2}- 0] \, dx = \int_{0}^{3} x \sqrt{25-x^2} \, dx \) which can be solved either using trigonometric substitution or by recognizing it as a form of a standard integral. Either way, after solving, the result is \(\frac{18}{5}\).