Question

If $M$ be the area of the region (in square units) defined by $\{(x, y) \in R^2: x^2 + y^2 \leq 5, y^2 \leq 4x, x \geq 1\}$, then $M$ is equal to

• A. $\frac{5\pi}{2} + 2 - 5\sin^{-1}(\frac{1}{\sqrt{5}})$
• B. $\frac{5\pi}{2} - 2 - 5\sin^{-1}(\frac{1}{\sqrt{5}})$
• C. $\frac{5\pi}{4} - 1 - \frac{5}{2}\sin^{-1}(\frac{1}{\sqrt{5}})$
• D. $\frac{5\pi}{4} + 1 - \frac{5}{2}\sin^{-1}(\frac{1}{\sqrt{5}})$

Answer 1

Answer: (B)

$\frac{5\pi}{2} - 2 - 5\sin^{-1}(\frac{1}{\sqrt{5}})$

Answer 2

The problem asks for the area of the region defined by the inequalities:

1. $x^2 + y^2 \leq 5$ (Region inside or on the circle centered at the origin with radius $R=\sqrt{5}$)
2. $y^2 \leq 4x$ (Region inside or on the parabola $y^2 = 4x$ opening to the right)
3. $x \geq 1$ (Region to the right of or on the vertical line $x=1$)

Let's find the intersection points of the boundary curves:

• Circle $x^2+y^2=5$ and Line $x=1$: Substitute $x=1$ into the circle equation: $1^2+y^2=5 \implies y^2=4 \implies y=\pm 2$. So the intersection points are $(1,2)$ and $(1,-2)$.
• Parabola $y^2=4x$ and Line $x=1$: Substitute $x=1$ into the parabola equation: $y^2=4(1) \implies y^2=4 \implies y=\pm 2$. So the intersection points are $(1,2)$ and $(1,-2)$.
• Circle $x^2+y^2=5$ and Parabola $y^2=4x$: Substitute $y^2=4x$ into the circle equation: $x^2+4x=5 \implies x^2+4x-5=0 \implies (x+5)(x-1)=0$. This gives $x=1$ or $x=-5$. Since $y^2=4x$, $x$ must be non-negative for real $y$. So $x=1$. If $x=1$, $y^2=4$, so $y=\pm 2$. The intersection points are $(1,2)$ and $(1,-2)$.

All three boundaries intersect at $(1,2)$ and $(1,-2)$.

Now, let's determine the actual region. We need the intersection of the three regions defined by the inequalities. Consider the region defined by $x^2+y^2 \leq 5$ and $x \geq 1$. This is a circular segment of the circle $x^2+y^2=5$ to the right of the line $x=1$. For this region, the $x$-values range from $1$ to $\sqrt{5}$ (the maximum $x$ for the circle).

Let's check if the condition $y^2 \leq 4x$ is redundant for points in this circular segment. For any point $(x,y)$ such that $x^2+y^2 \leq 5$ and $x \geq 1$: We know $y^2 \leq 5-x^2$. We need to check if $5-x^2 \leq 4x$ for $x \in [1, \sqrt{5}]$. The inequality $5-x^2 \leq 4x$ can be rewritten as $x^2+4x-5 \geq 0$. Factoring the quadratic, we get $(x+5)(x-1) \geq 0$. Since $x \geq 1$, $(x+5)$ is positive and $(x-1)$ is non-negative. Thus, their product $(x+5)(x-1)$ is always non-negative for $x \geq 1$. This means that for any point $(x,y)$ satisfying $x^2+y^2 \leq 5$ and $x \geq 1$, the condition $y^2 \leq 4x$ is automatically satisfied.

Therefore, the region $M$ is simply the area of the circular segment defined by $x^2+y^2 \leq 5$ and $x \geq 1$. The region is symmetric about the x-axis, so we can calculate the area for $y \geq 0$ and multiply by 2. For $y \geq 0$, the upper boundary is $y = \sqrt{5-x^2}$. The area $M$ is given by the integral:

$M = \int_{1}^{\sqrt{5}} 2\sqrt{5-x^2} dx$

This is a standard integral of the form $\int \sqrt{a^2-x^2} dx = \frac{x}{2}\sqrt{a^2-x^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right)$. Here $a^2=5$, so $a=\sqrt{5}$.

$M = 2 \left[ \frac{x}{2}\sqrt{5-x^2} + \frac{5}{2}\sin^{-1}\left(\frac{x}{\sqrt{5}}\right) \right]_{1}^{\sqrt{5}}$ $M = \left[ x\sqrt{5-x^2} + 5\sin^{-1}\left(\frac{x}{\sqrt{5}}\right) \right]_{1}^{\sqrt{5}}$

Now, evaluate the definite integral: At the upper limit $x=\sqrt{5}$:

$\sqrt{5}\sqrt{5-(\sqrt{5})^2} + 5\sin^{-1}\left(\frac{\sqrt{5}}{\sqrt{5}}\right) = \sqrt{5}\sqrt{0} + 5\sin^{-1}(1) = 0 + 5\left(\frac{\pi}{2}\right) = \frac{5\pi}{2}$

At the lower limit $x=1$:

$1\sqrt{5-1^2} + 5\sin^{-1}\left(\frac{1}{\sqrt{5}}\right) = \sqrt{4} + 5\sin^{-1}\left(\frac{1}{\sqrt{5}}\right) = 2 + 5\sin^{-1}\left(\frac{1}{\sqrt{5}}\right)$

Subtract the lower limit value from the upper limit value:

$M = \frac{5\pi}{2} - \left( 2 + 5\sin^{-1}\left(\frac{1}{\sqrt{5}}\right) \right)$ $M = \frac{5\pi}{2} - 2 - 5\sin^{-1}\left(\frac{1}{\sqrt{5}}\right)$