Calculate Area Bounded By Curves Y=√x, Y=√(8-x), And Y=0

In the realm of calculus, determining the area enclosed by curves is a fundamental concept with vast applications in various fields, including physics, engineering, and economics. This comprehensive guide delves into the process of calculating the area bounded by the functions $y = \sqrt{x}$, $y = \sqrt{8-x}$, and $y = 0$. We will explore the underlying principles, step-by-step calculations, and graphical interpretations to provide a thorough understanding of the topic.

Understanding the Problem

Before we embark on the solution, let's first understand the problem at hand. We are tasked with finding the area of the region enclosed by three curves:

• $y = \sqrt{x}$: This represents the upper half of a parabola opening to the right.
• $y = \sqrt{8-x}$: This represents the upper half of a parabola opening to the left.
• $y = 0$: This is the x-axis.

Visualizing these curves will help us grasp the shape of the region we're dealing with. The region is bounded by the x-axis from below and by the two square root functions from above. To accurately calculate the area, we need to find the points where these curves intersect.

To get started, the key here is recognizing the intersection points between these curves. These points define the boundaries of our region and are crucial for setting up the integrals. We need to find where $y = \sqrt{x}$ intersects $y = \sqrt{8-x}$ and where each of these curves intersects $y = 0$. These intersections will provide the limits of integration for calculating the enclosed area. Understanding these limits and the functions that bound the area is fundamental to solving this problem accurately. Additionally, a visual representation of the curves can significantly aid in comprehending the problem and verifying the solution.

Step 1: Finding the Intersection Points

To determine the area enclosed by the given functions, the first crucial step is to identify the points where these functions intersect. These intersection points define the boundaries of the region whose area we aim to calculate.

Intersection of $y = \sqrt{x}$ and $y = \sqrt{8-x}$

To find the intersection of these two curves, we set them equal to each other:

$\sqrt{x} = \sqrt{8-x}$

Squaring both sides to eliminate the square roots, we get:

$x = 8 - x$

Adding $x$ to both sides gives:

$2x = 8$

Dividing by 2, we find:

$x = 4$

Now, substitute this value of $x$ back into either equation to find the corresponding $y$ value. Using $y = \sqrt{x}$:

$y = \sqrt{4} = 2$

Therefore, the intersection point of $y = \sqrt{x}$ and $y = \sqrt{8-x}$ is $(4, 2)$. This point is vital as it represents where the two curves meet and helps delineate the region of interest.

Intersection with $y = 0$

Next, we find where each curve intersects the x-axis ($y = 0$):

• For $y = \sqrt{x}$, setting $y = 0$ gives:
  $0 = \sqrt{x}$
  Squaring both sides, we get $x = 0$. So, the intersection point is $(0, 0)$.
• For $y = \sqrt{8-x}$, setting $y = 0$ gives:
  $0 = \sqrt{8-x}$ Squaring both sides, we get $0 = 8 - x$, which implies $x = 8$. So, the intersection point is $(8, 0)$.

These intersection points, $(0, 0)$ and $(8, 0)$, indicate where the curves meet the x-axis, further defining the boundaries of the area we want to compute. Identifying these points is a critical step, as they will serve as the limits of integration in our subsequent calculations. Understanding the significance of these intersections is paramount to setting up the integral correctly and accurately determining the area.

Step 2: Setting Up the Integrals

With the intersection points determined, we can now formulate the integrals required to calculate the area. The region is bounded by the curves $y = \sqrt{x}$, $y = \sqrt{8-x}$, and $y = 0$. We've found the intersection points to be $(0, 0)$, $(4, 2)$, and $(8, 0)$. This tells us the region is divided into two parts, each requiring a separate integral.

Region 1: From $x = 0$ to $x = 4$

In this region, the area is bounded above by the curve $y = \sqrt{x}$ and below by the x-axis ($y = 0$). Therefore, the integral for this region is:

$\int_{0}^{4} \sqrt{x} \, dx$

This integral represents the area under the curve $y = \sqrt{x}$ from $x = 0$ to $x = 4$. It captures the portion of the area where the square root of x defines the upper boundary.

Region 2: From $x = 4$ to $x = 8$

In this region, the area is bounded above by the curve $y = \sqrt{8-x}$ and below by the x-axis ($y = 0$). Thus, the integral for this region is:

$\int_{4}^{8} \sqrt{8-x} \, dx$

This integral represents the area under the curve $y = \sqrt{8-x}$ from $x = 4$ to $x = 8$. It covers the part of the area where the square root of 8 minus x forms the upper boundary.

The total area is the sum of these two integrals. By setting up these integrals correctly, we've effectively divided the problem into manageable parts. The next step involves evaluating these integrals, which will give us the numerical values for each area. Accurately setting up these integrals is pivotal for the final result; any error here will propagate through the rest of the calculation. Understanding the geometry of the region and how the curves define its boundaries is essential for this step.

Step 3: Evaluating the Integrals

Now that we have set up the integrals, the next step is to evaluate them to find the areas of the two regions. This involves applying the fundamental theorem of calculus and finding the antiderivatives of the functions.

Evaluating $\int_{0}^{4} \sqrt{x} \, dx$

First, we rewrite $\sqrt{x}$ as $x^{\frac{1}{2}}$. The antiderivative of $x^{\frac{1}{2}}$ is $\frac{2}{3}x^{\frac{3}{2}}$. Now, we evaluate this antiderivative at the limits of integration:

$\left[ \frac{2}{3}x^{\frac{3}{2}} \right]_{0}^{4} = \frac{2}{3}(4^{\frac{3}{2}}) - \frac{2}{3}(0^{\frac{3}{2}}) = \frac{2}{3}(8) - 0 = \frac{16}{3}$

So, the area of the first region is $\frac{16}{3}$ square units.

Evaluating $\int_{4}^{8} \sqrt{8-x} \, dx$

For this integral, we use a u-substitution. Let $u = 8 - x$, so $du = -dx$. When $x = 4$, $u = 4$, and when $x = 8$, $u = 0$. The integral becomes:

$-\int_{4}^{0} \sqrt{u} \, du$

We can reverse the limits of integration and change the sign:

$\int_{0}^{4} \sqrt{u} \, du$

This is the same integral we evaluated before, just with a different variable. The antiderivative of $\sqrt{u}$ is $\frac{2}{3}u^{\frac{3}{2}}$. Evaluating at the limits:

$\left[ \frac{2}{3}u^{\frac{3}{2}} \right]_{0}^{4} = \frac{2}{3}(4^{\frac{3}{2}}) - \frac{2}{3}(0^{\frac{3}{2}}) = \frac{2}{3}(8) - 0 = \frac{16}{3}$

Thus, the area of the second region is also $\frac{16}{3}$ square units.

Evaluating these integrals is a crucial step, as it converts the symbolic representation of the area into numerical values. The application of the fundamental theorem of calculus and techniques like u-substitution are essential skills in this process. Carefully executing these steps ensures an accurate final answer.

Step 4: Summing the Areas and Rounding

Having evaluated the individual integrals, we now sum the areas of the two regions to obtain the total area bounded by the curves. This final calculation provides the answer to our original problem.

We found that the area of the first region, bounded by $y = \sqrt{x}$ and $y = 0$ from $x = 0$ to $x = 4$, is $\frac{16}{3}$ square units. The area of the second region, bounded by $y = \sqrt{8-x}$ and $y = 0$ from $x = 4$ to $x = 8$, is also $\frac{16}{3}$ square units.

To find the total area, we add these two areas together:

$\text{Total Area} = \frac{16}{3} + \frac{16}{3} = \frac{32}{3}$

Now, we convert this fraction to a decimal to round it to the nearest thousandth:

$\frac{32}{3} \approx 10.66666...$

Rounding to the nearest thousandth, we get:

$\text{Total Area} \approx 10.667$

Therefore, the total area bounded by the functions $y = \sqrt{x}$, $y = \sqrt{8-x}$, and $y = 0$ is approximately 10.667 square units. This final calculation provides the quantitative answer we sought. Accuracy in summing the areas and rounding appropriately is essential to presenting the correct result. This completes the process of finding the area bounded by the given curves, demonstrating the application of calculus principles to solve a geometric problem.

Conclusion

In conclusion, we have successfully determined the area bounded by the curves $y = \sqrt{x}$, $y = \sqrt{8-x}$, and $y = 0$. By following a systematic approach involving finding intersection points, setting up integrals, evaluating them, and summing the results, we arrived at the solution of approximately 10.667 square units.

This process highlights the power of calculus in solving geometric problems. The steps we took demonstrate a general methodology applicable to finding areas bounded by various functions. From identifying intersection points to setting up and evaluating integrals, each step is crucial in obtaining an accurate result. This problem serves as a valuable example for understanding and applying the concepts of definite integrals and area calculation.

The ability to find areas bounded by curves has significant practical applications in various fields, from engineering design to economic modeling. The principles demonstrated here are fundamental to many advanced mathematical and scientific concepts. By mastering these techniques, one can tackle more complex problems involving area, volume, and other geometric quantities. Therefore, a thorough understanding of these concepts is essential for anyone pursuing studies or careers in STEM fields. The exercise of solving this problem not only provides a specific answer but also reinforces a broader skill set applicable to a wide range of mathematical challenges.