The area of the region $\{(x,y):\lvert x-y\rvert \le y \le 4\sqrt{x}\}$ is:

Step 1: Break down the inequalities

The given region is defined by two inequalities: $\lvert x-y\rvert \le y$ and $y \le 4\sqrt{x}$. We first break down the absolute value inequality. The inequality $\lvert x-y\rvert \le y$ can be split into two separate inequalities: $-(x-y) \le y$ and $y \le x-y$.

Step 2: Simplify the inequalities

From $-(x-y) \le y$, we get $-x+y \le y$, which simplifies to $-x \le 0$, or $x \ge 0$. This is consistent with $y \le 4\sqrt{x}$, which requires $x \ge 0$. From $y \le x-y$, we get $2y \le x$, which simplifies to $y \le \frac{x}{2}$. However, the first part of the absolute value inequality $-(x-y) \le y$ actually gives $y \ge x/2$. Let's re-evaluate. $-(x-y) \le y \implies -x+y \le y \implies -x \le 0 \implies x \ge 0$. This is not correct. Let's re-evaluate the absolute value inequality. $\lvert x-y\rvert \le y$ means $-y \le x-y \le y$. The left part is $-y \le x-y \implies 0 \le x$, which is $x \ge 0$. The right part is $x-y \le y \implies x \le 2y \implies y \ge \frac{x}{2}$. So the first inequality simplifies to $y \ge \frac{x}{2}$ (and $x \ge 0$). The second inequality is $y \le 4\sqrt{x}$.

Step 3: Find intersection points

We need to find the intersection points of the two boundary curves $y = \frac{x}{2}$ and $y = 4\sqrt{x}$. Set the expressions for $y$ equal to each other: $\frac{x}{2} = 4\sqrt{x}$. Squaring both sides gives $\frac{x^2}{4} = 16x$. Multiplying by 4 gives $x^2 = 64x$. Rearranging, we get $x^2 - 64x = 0$, which factors as $x(x-64) = 0$. This gives $x=0$ and $x=64$.

Step 4: Calculate the area using integration

The area of the region is given by the definite integral of the upper curve minus the lower curve, from the first intersection point to the second. In this case, $y = 4\sqrt{x}$ is the upper curve and $y = \frac{x}{2}$ is the lower curve between $x=0$ and $x=64$.

Step 5: Evaluate the integral

Now we evaluate the definite integral. We integrate term by term: $\int 4x^{1/2} dx = 4 \cdot \frac{x^{3/2}}{3/2} = \frac{8}{3}x^{3/2}$ and $\int \frac{x}{2} dx = \frac{x^2}{4}$. Then we apply the limits of integration.

Step 6: Calculate the final area

Substitute the upper limit $x=64$ and the lower limit $x=0$ into the antiderivative. $(64)^{3/2} = (\sqrt{64})^3 = 8^3 = 512$. $(64)^2 = 4096$. Simplify the expression to find the final area.

More from this chapter

• Find the area bounded by the curves y = x³ and y = 4x.
• (a) Find the consumer's surplus for the demand function p = 25 - x - x², where the prevailing market price p_0 = 19. OR (b) Solve the following initial value di
• Find the length of the curve defined parametrically by x = a(cos t + t sin t), y = a(sin t - t cos t) for 0 ≤ t ≤ π/2.
• The area of the region, inside the circle (x - 2√3)² + y² = 12 and outside the parabola y² = 2√3 x is:
• Find the area of the region bounded by the curves y = x² and y = |x|.
• A cylindrical tank with radius 2 m and height 5 m is filled with water. Find the work required to pump all the water to the top of the tank.
• Find the area of the region bounded by the curves y² = 4x and x² = 4y.
• Find the arc length of the curve y = ln(sec x) from x = 0 to x = π/4.

All Application of Integrals solutions →