Question

Consider the region $R=\{(x, y): x^2 + y^2 \leq 100, \sin(x+y) > 0\}$. What is the area of $R$?

• A. $25\pi$
• B. $50\pi$
• C. 50
• D. $100\pi - 50$

Answer 1

Answer: (B)

50π

Answer 2

The problem asks for the area of the region $R=\{(x, y): x^2 + y^2 \leq 100, \sin(x+y) > 0\}$.

Let's analyze the two conditions defining the region $R$:

1. $x^2 + y^2 \leq 100$: This inequality describes a closed disk centered at the origin $(0,0)$ with a radius of $r = \sqrt{100} = 10$. Let's call this disk $D$. The total area of this disk is $A_D = \pi r^2 = \pi (10)^2 = 100\pi$.

2. $\sin(x+y) > 0$: This inequality specifies which part of the disk we are interested in. The sine function is positive when its argument lies in the intervals $(2n\pi, (2n+1)\pi)$ for any integer $n$. So, we need $x+y \in (2n\pi, (2n+1)\pi)$. This means the region $R$ consists of parts of the disk $D$ that fall within these parallel strips defined by $2n\pi < x+y < (2n+1)\pi$.

To find the area of $R$, we can use a symmetry argument. Let $f(x,y) = \sin(x+y)$. The region $D$ (the disk $x^2+y^2 \leq 100$) is symmetric with respect to the origin. This means that if a point $(x,y)$ is in $D$, then its antipodal point $(-x,-y)$ is also in $D$.

Now let's examine the condition $\sin(x+y) > 0$: Consider a point $(x,y)$ in $R$. By definition, $(x,y) \in D$ and $\sin(x+y) > 0$. Now consider the antipodal point $(-x,-y)$. Since $(x,y) \in D$ and $D$ is symmetric about the origin, $(-x,-y)$ is also in $D$. Let's evaluate $\sin$ at this new point: $\sin((-x)+(-y)) = \sin(-(x+y))$ Using the trigonometric identity $\sin(-\theta) = -\sin(\theta)$, we get: $\sin(-(x+y)) = -\sin(x+y)$.

Since we know that $\sin(x+y) > 0$ for $(x,y) \in R$, it follows that for the antipodal point $(-x,-y)$, $\sin((-x)+(-y)) = -\sin(x+y) < 0$.

Let $R_+$ be the region where $\sin(x+y) > 0$ within the disk $D$. So $R = R_+$. Let $R_-$ be the region where $\sin(x+y) < 0$ within the disk $D$. Let $R_0$ be the region where $\sin(x+y) = 0$ within the disk $D$. This corresponds to lines $x+y=k\pi$ for integers $k$. The area of a finite set of lines is zero. Therefore, the total area of the disk $D$ is $Area(D) = Area(R_+) + Area(R_-)$.

The mapping $T(x,y) = (-x,-y)$ is a rotation by $180^\circ$ about the origin. This transformation preserves area. We showed that if $(x,y) \in R_+$, then $T(x,y) = (-x,-y) \in R_-$. Similarly, if $(x,y) \in R_-$, then $T(x,y) = (-x,-y) \in R_+$. This establishes a one-to-one, area-preserving correspondence between the points in $R_+$ and the points in $R_-$. Therefore, $Area(R_+) = Area(R_-)$.

Since $Area(D) = Area(R_+) + Area(R_-)$ and $Area(R_+) = Area(R_-)$, we have: $Area(D) = 2 \times Area(R_+)$ $100\pi = 2 \times Area(R_+)$ $Area(R_+) = \frac{100\pi}{2} = 50\pi$.

The area of region $R$ is $50\pi$.