Question

for this problem type the word theta for \\( \theta \\). the region \\( r \\) is bounded by the \\( x \\)-axis, the straight line in the graph, and the vertical line \\( x = 2 \\). then
\\ \iint _ { r } x ^ { 2 } + y ^ { 2 } d a = \int \quad \int \quad \int \quad d r d \theta \\

Explanation:

Step1: Determine the range of \(\theta\)

The line in the graph has a slope \(m = - 1\), so the angle \(\theta\) from the positive \(x\) - axis is \(\theta=\frac{7\pi}{4}\) (since it is in the fourth quadrant). The region is bounded by the \(x\) - axis (\(\theta = 0\)) and the line \(y=-x + 2\) (converted to polar form \(r\sin\theta=-r\cos\theta + 2\Rightarrow r=\frac{2}{\sin\theta+\cos\theta}\)). So, \(\theta\) ranges from \(0\) to \(\frac{3\pi}{4}\).

Step2: Determine the range of \(r\)

The lower limit of \(r\) is \(r = 0\) (origin). The upper limit of \(r\) is found from the line \(x + y=2\). In polar coordinates \(x=r\cos\theta\), \(y = r\sin\theta\), so \(r\cos\theta+r\sin\theta=2\Rightarrow r=\frac{2}{\cos\theta+\sin\theta}\)

Step3: Set up the double - integral in polar coordinates

We know that \(x^{2}+y^{2}=r^{2}\) and \(dA = r\space dr\space d\theta\). The double - integral \(\iint_{R}(x^{2}+y^{2})dA\) in polar coordinates is \(\int_{0}^{\frac{3\pi}{4}}\int_{0}^{\frac{2}{\cos\theta+\sin\theta}}r^{2}\cdot r\space dr\space d\theta=\int_{0}^{\frac{3\pi}{4}}\int_{0}^{\frac{2}{\cos\theta+\sin\theta}}r^{3}\space dr\space d\theta\)

Answer:

\(\int_{0}^{\frac{3\pi}{4}}\int_{0}^{\frac{2}{\cos\theta+\sin\theta}}r^{3}\space dr\space d\theta\)