Centroid Calculation Find The Centroid Of Region Bounded By Y=4x^2 And Y=x^2+6

Hey guys! Today, we're diving into a fun problem: finding the centroid of a region tucked between two curves. It might sound a bit intimidating, but trust me, we'll break it down step by step. We're dealing with the functions $y = 4x^2$ and $y = x^2 + 6$, and our mission is to pinpoint the centroid, that magical point $(\bar{x}, \bar{y})$ that represents the geometric center of the area between these curves. Let's get started!

Understanding the Centroid

Before we jump into calculations, let's make sure we're all on the same page about what a centroid actually is. Imagine you have a flat shape cut out of cardboard. The centroid is the point where you could perfectly balance that shape on the tip of a pin. It's essentially the average position of all the points within the shape. In mathematical terms, the centroid $(\bar{x}, \bar{y})$ of a region is calculated using integrals, which help us find the "average" x and y coordinates over the entire region. This concept has many applications in physics and engineering, where understanding the center of mass or balance is crucial.

Formulas for the Centroid

Okay, let's get a little technical. To find the centroid, we'll use these formulas:

• $\bar{x} = \frac{1}{A} \int_{a}^{b} x[f(x) - g(x)] dx$
• $\bar{y} = \frac{1}{A} \int_{a}^{b} \frac{1}{2} [f(x)^2 - g(x)^2] dx$

Where:

• $A$ is the area of the region.
• $f(x)$ is the upper function (the one with the larger y-values).
• $g(x)$ is the lower function (the one with the smaller y-values).
• $a$ and $b$ are the x-coordinates of the points where the curves intersect.

These formulas might look a bit scary, but don't worry! We'll take them apart piece by piece. The key thing to remember is that we need to find the area $A$, identify the upper and lower functions, and determine the limits of integration $a$ and $b$.

Step 1: Finding the Intersection Points

First things first, we need to figure out where the curves $y = 4x^2$ and $y = x^2 + 6$ intersect. These intersection points will give us the limits of integration, $a$ and $b$. To find them, we simply set the two equations equal to each other:

$4x^2 = x^2 + 6$

Now, let's solve for $x$:

$3x^2 = 6$ $x^2 = 2$ $x = \pm \sqrt{2}$

So, our curves intersect at $x = -\sqrt{2}$ and $x = \sqrt{2}$. These are our limits of integration, $a = -\sqrt{2}$ and $b = \sqrt{2}$. Great job! We've got the first piece of the puzzle.

Visualizing the Region

It's always a good idea to visualize what we're working with. Imagine the two parabolas, $y = 4x^2$ (which is narrower) and $y = x^2 + 6$ (which is wider and shifted upwards). The region we're interested in is the area enclosed between these two curves. Visualizing the region helps us confirm which function is on top (the upper function) and which is on the bottom (the lower function).

In this case, $y = x^2 + 6$ is the upper function, and $y = 4x^2$ is the lower function within the interval $[-\sqrt{2}, \sqrt{2}]$. This understanding is crucial for setting up our integrals correctly.

Step 2: Calculating the Area (A)

Next up, we need to find the area ($A$) of the region between the curves. We'll use the following integral:

$A = \int_{a}^{b} [f(x) - g(x)] dx$

In our case, $f(x) = x^2 + 6$, $g(x) = 4x^2$, $a = -\sqrt{2}$, and $b = \sqrt{2}$. So, let's plug those values in:

$A = \int_{-\sqrt{2}}^{\sqrt{2}} [(x^2 + 6) - 4x^2] dx$

Simplify the integrand:

$A = \int_{-\sqrt{2}}^{\sqrt{2}} [6 - 3x^2] dx$

Now, we integrate:

$A = [6x - x^3]_{-\sqrt{2}}^{\sqrt{2}}$

Evaluate at the limits of integration:

$A = (6\sqrt{2} - (\sqrt{2})^3) - (6(-\sqrt{2}) - (-\sqrt{2})^3)$ $A = (6\sqrt{2} - 2\sqrt{2}) - (-6\sqrt{2} + 2\sqrt{2})$ $A = 4\sqrt{2} + 4\sqrt{2}$ $A = 8\sqrt{2}$

So, the area of the region is $8\sqrt{2}$. Awesome! We're making great progress.

Why Area Matters

Calculating the area is a fundamental step because it appears in the denominators of our centroid formulas. Think of it this way: the centroid's coordinates are a weighted average, and the area acts as a normalization factor, ensuring that the centroid's coordinates are within the bounds of the region. Without the area, our centroid calculations would be off.

Step 3: Calculating $\bar{x}$

Now, let's find the x-coordinate of the centroid, $\bar{x}$. Remember the formula:

$\bar{x} = \frac{1}{A} \int_{a}^{b} x[f(x) - g(x)] dx$

We already know $A = 8\sqrt{2}$, $f(x) = x^2 + 6$, $g(x) = 4x^2$, $a = -\sqrt{2}$, and $b = \sqrt{2}$. Let's plug those in:

$\bar{x} = \frac{1}{8\sqrt{2}} \int_{-\sqrt{2}}^{\sqrt{2}} x[(x^2 + 6) - 4x^2] dx$

Simplify the integrand:

$\bar{x} = \frac{1}{8\sqrt{2}} \int_{-\sqrt{2}}^{\sqrt{2}} x[6 - 3x^2] dx$ $\bar{x} = \frac{1}{8\sqrt{2}} \int_{-\sqrt{2}}^{\sqrt{2}} [6x - 3x^3] dx$

Now, integrate:

$\bar{x} = \frac{1}{8\sqrt{2}} [3x^2 - \frac{3}{4}x^4]_{-\sqrt{2}}^{\sqrt{2}}$

Evaluate at the limits of integration:

$\bar{x} = \frac{1}{8\sqrt{2}} [(3(\sqrt{2})^2 - \frac{3}{4}(\sqrt{2})^4) - (3(-\sqrt{2})^2 - \frac{3}{4}(-\sqrt{2})^4)]$ $\bar{x} = \frac{1}{8\sqrt{2}} [(6 - 3) - (6 - 3)]$ $\bar{x} = \frac{1}{8\sqrt{2}} [3 - 3]$ $\bar{x} = 0$

So, $\bar{x} = 0$. Fantastic! The x-coordinate of the centroid is 0. This makes sense, right? Because the region is symmetric about the y-axis.

Symmetry and Centroids

The fact that $\bar{x} = 0$ highlights an important concept: symmetry. If a region is symmetric about a vertical line, the centroid will lie on that line. This can often save us calculation time, as we can immediately deduce one of the centroid's coordinates.

Step 4: Calculating $\bar{y}$

Finally, let's calculate the y-coordinate of the centroid, $\bar{y}$. Here's the formula:

$\bar{y} = \frac{1}{A} \int_{a}^{b} \frac{1}{2} [f(x)^2 - g(x)^2] dx$

Again, we know $A = 8\sqrt{2}$, $f(x) = x^2 + 6$, $g(x) = 4x^2$, $a = -\sqrt{2}$, and $b = \sqrt{2}$. Let's plug in:

$\bar{y} = \frac{1}{8\sqrt{2}} \int_{-\sqrt{2}}^{\sqrt{2}} \frac{1}{2} [(x^2 + 6)^2 - (4x^2)^2] dx$

Simplify the integrand:

$\bar{y} = \frac{1}{16\sqrt{2}} \int_{-\sqrt{2}}^{\sqrt{2}} [(x^4 + 12x^2 + 36) - 16x^4] dx$ $\bar{y} = \frac{1}{16\sqrt{2}} \int_{-\sqrt{2}}^{\sqrt{2}} [-15x^4 + 12x^2 + 36] dx$

Now, integrate:

$\bar{y} = \frac{1}{16\sqrt{2}} [-3x^5 + 4x^3 + 36x]_{-\sqrt{2}}^{\sqrt{2}}$

Evaluate at the limits of integration:

$\bar{y} = \frac{1}{16\sqrt{2}} [(-3(\sqrt{2})^5 + 4(\sqrt{2})^3 + 36\sqrt{2}) - (-3(-\sqrt{2})^5 + 4(-\sqrt{2})^3 + 36(-\sqrt{2}))]$ $\bar{y} = \frac{1}{16\sqrt{2}} [(-12\sqrt{2} + 8\sqrt{2} + 36\sqrt{2}) - (12\sqrt{2} - 8\sqrt{2} - 36\sqrt{2})]$ $\bar{y} = \frac{1}{16\sqrt{2}} [32\sqrt{2} - (-32\sqrt{2})]$ $\bar{y} = \frac{1}{16\sqrt{2}} [64\sqrt{2}]$ $\bar{y} = 4$

So, $\bar{y} = 4$. Excellent! We've found the y-coordinate of the centroid.

The Grand Finale: The Centroid

We've done it, guys! We've successfully calculated the centroid of the region bounded by the curves $y = 4x^2$ and $y = x^2 + 6$. The centroid is located at:

$(\bar{x}, \bar{y}) = (0, 4)$

Checking Our Work

It's always a good idea to take a moment and think about whether our answer makes sense. We found that the centroid is at (0, 4). This point lies on the y-axis, which aligns with the symmetry of the region. Also, the y-coordinate of 4 seems reasonable given the shape and position of the region between the two parabolas.

Final Thoughts

Finding the centroid involves a few key steps: identifying the region, finding the intersection points, calculating the area, and then using the centroid formulas. It's a process that combines algebra, calculus, and a bit of spatial reasoning. But hey, you nailed it! Keep practicing, and you'll become a centroid-calculating pro in no time. You rock!