Question

Find the centroid of the region bounded by the given curves. Make a sketch and use symmetry where possible. $$ x=y^{2}, x=2 $$

Short answer

The centroid is at (1, 0).

Step-by-step solution

Understand the Region

The curves given are:1. \( x = y^2 \), which is a parabola opening to the right with its vertex at the origin 2. \( x = 2 \), which is a vertical line. We need to find the bounded region between these two curves. The parabola \( x = y^2 \) intersects the vertical line \( x = 2 \) at points where \( y^2 = 2 \), giving the intersections \( y = \sqrt{2} \) and \( y = -\sqrt{2} \). The region of interest is between these y-values.

Set Up the Integral Formula for Area

The area \( A \) of the bounded region is given by:\[ A = \int_{-\sqrt{2}}^{\sqrt{2}} (2 - y^2) \, dy \] This integral finds the area between the parabola \( x = y^2 \) and the line \( x = 2 \).

Calculate the Area

Solve the integral to find the area:\[ A = \int_{-\sqrt{2}}^{\sqrt{2}} (2 - y^2) \, dy = \left[ 2y - \frac{y^3}{3} \right]_{-\sqrt{2}}^{\sqrt{2}} \]Calculate and simplify:\[ = \left( 2\sqrt{2} - \frac{(\sqrt{2})^3}{3} \right) - \left( 2(-\sqrt{2}) + \frac{(-\sqrt{2})^3}{3} \right) \]\[ = 4\sqrt{2} - \frac{4\sqrt{2}}{3} = \frac{8\sqrt{2}}{3} \]

Find the Centroid Coordinates

The centroid \( (\bar{x}, \bar{y}) \) has coordinates given by\[ \bar{x} = \frac{1}{A} \int_{-\sqrt{2}}^{\sqrt{2}} x_{mid} \cdot (2 - y^2) \, dy \]For a region symmetric about the y-axis, \( \bar{y} = 0 \). Here, \( x_{mid} = \frac{y^2 + 2}{2} \).

Calculate \( \bar{x} \)

The coordinates \( \bar{x} \) of the centroid is:\[ \bar{x} = \frac{1}{\frac{8\sqrt{2}}{3}} \int_{-\sqrt{2}}^{\sqrt{2}} \left( \frac{y^2 + 2}{2} \right) (2-y^2) \, dy \]This simplifies to:\[ \bar{x} = \frac{3}{8\sqrt{2}} \int_{-\sqrt{2}}^{\sqrt{2}} \left( 2 - y^2 \right) \, dy \]\[ \bar{x} = 1 \]

Compile Centroid

Combine \( \bar{x} \) and \( \bar{y} \) to find the centroid:Since the region is symmetric about the x-axis, \( \bar{y} = 0 \). So, the centroid is at \( (1, 0) \).

Key concepts

Integrals in Calculus

When finding the centroid of a region bound by curves, integrals play a crucial role. The integral helps us compute quantities like area and boundary values, which are essential for determining centroidal coordinates.
In our exercise, the integral was used to calculate the area under the curves, which is the space between the parabolic curve and the vertical line. This area, denoted as \( A \), is crucial for finding the centroid because it appears in the denominator when calculating the centroid coordinates.
We set up the integral as:

• To find the area \( A \) between the parabola \( x = y^2 \) and the line \( x = 2 \), we integrated the function \( 2 - y^2 \) over the interval \([-\sqrt{2}, \sqrt{2}]\) which signifies the bounds of the region.
• The solution involved integrating this function to obtain the area \( A = \frac{8\sqrt{2}}{3} \).
• Then, this area is used in the next step to find \( \bar{x} \), the x-coordinate of the centroid.

Integrals help simplify these calculations by systematically accounting for the varying shapes and spaces between mathematical curves. Without integrals, calculating centroids for irregular regions would be much more difficult and less precise.

Symmetry in Mathematics

Symmetry simplifies many mathematical processes, including finding centroids. In our problem, symmetry significantly reduced the complexity of calculations.
The given region is formed by the parabola \( x = y^2 \), which is symmetric about the x-axis, and the vertical line \( x = 2 \). This symmetry allows us to infer certain properties without detailed computation:

• Because the parabola is symmetric about the y-axis, this also implies that the entire region is symmetric, allowing us to immediately deduce that the centroid \( \bar{y} \) is \( 0 \).
• Finding \( \bar{y} = 0 \) saves a step, focusing our calculation solely on the x-coordinate, \( \bar{x} \).
• Symmetry ensures that the region's balance point horizontally lies along the x-axis.

Leveraging symmetry not only simplifies calculations, it also provides deeper insights into how geometric shapes behave, which is especially useful in areas like physics and engineering.

Parabolas and Lines Intersection

The intersection points between parabolas and lines are fundamental to determining the boundaries of regions to analyze further. In this exercise, the intersection of the parabola \( x = y^2 \) with the vertical line \( x = 2 \) is the starting point for analysing the region's shape.
We calculate the intersections to clearly define the bounds:

• The parabola \( x = y^2 \) opens to the right, while the line \( x = 2 \) is vertical, meaning they intersect where \( y^2 = 2 \).
• This results in the points \( y = \sqrt{2} \) and \( y = -\sqrt{2} \). These intersections allow us to calculate the integral over the interval \([-\sqrt{2}, \sqrt{2}]\), which are the y-bounds of the region.
• Understanding where these curves intersect helps in visualizing the shape of the bounded region, critical for setting up integrals and solving for regions' characteristics such as area and centroid.

Being able to identify intersections accurately is crucial for defining the portion of the graph where the region of interest lies.