This question aims to find the centroid of the region which is bounded by curves in the first quadrant.
A centroid is the center point of any shape or object and in this case the center point of any shape drawn in 2D. Another way to define the Centroid is, the point of the region where the region is balanced horizontally when suspended from that point.
The region defined in this question lies in the first quadrant of the cartesian plane which means that the values of $x-axis$ and $y-axis$ points are positive. The region is formed by the two curves that intersect each other at two different points in the first quadrant.
First, we will find the area, $A$, of the region between the points of intersection of two curves, and then we will find the Centroid by computing the moments. Moments of any region measure the tendency of that region to rotate about the origin. The Centroid $C$ will be:
\[ C = \left( \dfrac{M_y}{A}, \dfrac{M_x}{A} \right) \]
where $M_x$ and $M_y$ are the $x$ and $y$ moments respectively.
Expert Answer:
As discussed above, the region formed by the two curves is shown in Figure 1.
We will find the centroid of the region by finding its area and its moments. There will be two moments for this region, $x$-moment, and $y$-moment. We divide $y$-moment by the area to get $x$-coordinate and divide the $x$-moment by the area to get $y$-coordinate.
The area, $A$, of the region can be found by:
\[ A = \int_{a}^{b} f(x) – g(x) \,dx \]
Here, $a$ and $b$ shows the limits of the region with respect to $x-axis$. $a$ is the lower limit and $b$ is the upper limit. Here
\[ [a, b] = [0, 1] \]
We have
\[ f(x) = x^3 \]
\[ g(x) = x^{1/3} \]
Substituting the values in the above equation, we get
\[ A = \int_{0}^{1} x^3 – x^{1/3} \,dx \]
Separating the integrations, we get
\[ A = \int_{0}^{1} x^3 \,dx – \int_{0}^{1} x^{1/3} \,dx \]
Solving separate integrations, we get
\[ A = \Big{[} \dfrac{x^4}{4} – \dfrac{3x^{4/3}}{4} \Big{]}_{0}^{1} \]
Substituting the upper and lower limits in the equation, we get
\[ A = \Big{[} \dfrac{1^4}{4} – \dfrac{3(1)^{4/3}}{4} \Big{]} – \Big{[} \dfrac{0^4}{4} – \dfrac{3(0)^{4/3}}{4} \Big{]} \]
After further we get,
\[ A = -0.5 \text{(units)$^2$} \]
Now we need to find the moments of the region.
$x$-moment is given by,
\[ M_x = \int_{a}^{b} \dfrac{1}{2} \{ (f(x))^2 – (g(x))^2 \} \,dx \]
Substituting the values,
\[ M_x = \int_{0}^{1} \dfrac{1}{2} \{ (x^3)^2 – (x^{1/3})^2 \} \,dx \]
Taking the constant out from integration,
\[ M_x = \dfrac{1}{2} \int_{0}^{1} x^6 – x^{2/3} \,dx \]
Separating the integrations,
\[ M_x = \dfrac{1}{2} \Big{[} \int_{0}^{1} x^6 \,dx – \int_{0}^{1} x^{2/3} \,dx \]
Solving the integrations,
\[ M_x = \dfrac{1}{2} \Big{[} \dfrac{x^7}{7} – \dfrac{3x^{5/3}}{5} \Big{]}_{0}^{1} \]
\[ M_x = \dfrac{1}{2} \bigg{[} \Big{[} \dfrac{1^7}{7} – \dfrac{3(1)^{5/3}}{5} \Big{]} – \Big{[} \dfrac{0^7}{7} – \dfrac{3(0)^{5/3}}{5} \Big{]} \bigg{]} \]
Simplifying,
\[ M_x = -0.23 \]
$y$-moment is given by,
\[ M_y = \int_{a}^{b} x \{ f(x) – g(x) \} \,dx \]
Substituting the values,
\[ M_y = \int_{0}^{1} x \{ x^3 – x^{1/3} \} \,dx \]
\[ M_y = \int_{0}^{1} x^4 – x^{5/3} \,dx \]
Separating the integrations,
\[ M_y = \int_{0}^{1} x^4 \,dx – \int_{0}^{1} x^{5/3} \} \,dx \]
Solving the integrations,
\[ M_y = \Big{[} \dfrac{x^5}{5} – \dfrac{3x^{8/3}}{8} \Big{]}_{0}^{1} \]
Substituting the limits,
\[ M_y = \Big{[}\Big{[} \dfrac{1^5}{5} – \dfrac{3(1)^{8/3}}{8} \Big{]} – \Big{[} \Big{[} \dfrac{0^5}{5} – \dfrac{3(0)^{8/3}}{8} \Big{]} \Big{]} \]
Simplifying,
\[ M_y = -0.23 \]
Numerical Results:
Let’s say the coordiantes of the Centroid of the region are: $( \overline{x} , \overline{y} )$. Using the area, $A$, the coordinates can be found as follows:
\[ \overline{x} = \dfrac{1}{A} \int_{a}^{b} x \{ f(x) -g(x) \} \,dx \]
Substituting values from above solved equations,
\[ \overline{x} = \dfrac{-0.23}{-0.5} \]
\[ \overline{x} = 0.46\]
And,
\[ \overline{y} = \dfrac{1}{A} \int_{a}^{b} \dfrac{1}{2} \{ (f(x))^2 – (g(x))^2 \} \,dx \]
Substituting values from above solved equations,
\[ \overline{y} = \dfrac{-0.23}{-0.5} \]
\[ \overline{y} = 0.46 \]
\[ ( \overline{x} , \overline{y} ) = (0.46, 0.46) \]
$( \overline{x} , \overline{y} )$ are the coordinates of the centroid of given region shown in Figure 1.
Alternative Solution:
When the values of moments of the region and area of the region are given. We can find the centroid values by directly substituting the values in following formulae.
\[ \overline{x} = \dfrac{M_y}{A} \]
\[ \overline{y} = \dfrac{M_x}{A} \]
Centroid Coordinates,
\[ ( \overline{x} , \overline{y} ) \]
Example:
Find the centroid of the region bounded by curves $y=x^4$ and $x=y^4$ on the interval $[0, 1]$ in the first quadrant shown in Figure 2.
Let,
\[ f(x) = x^4 \]
\[ g(x) = x^{1/4} \]
\[ [a, b] = [0, 1] \]
In this problem, we are given a smaller region from a shape formed by two curves in the first quadrant. It can also be solved by the method discussed above.
Area of the region in Figure 2 is given by,
\[ A = \int_{a}^{b} f(x) – g(x) \,dx \]
Substituting the values,
\[ A = \int_{0}^{1} x^4 – x^{1/4} \,dx \]
Solving the integration
\[ A = \Big{[} \dfrac{x^5}{5} – \dfrac{4x^{5/4}}{5} \Big{]}_{0}^{1} \]
Solving for limit values,
\[ A = \Big{[} \dfrac{1^5}{5} – \dfrac{4(1)^{5/4}}{5} \Big{]} – \Big{[} \dfrac{0^5}{5} – \dfrac{4(0)^{5/4}}{5} \Big{]} \]
Simplifying,
\[ A = -0.6 \text{(units)$^2$} \]
Now we find the moments of the region:
$x$-moment is given by,
\[ M_x = \int_{a}^{b} \dfrac{1}{2} \{ (f(x))^2 – (g(x))^2 \} \,dx \]
Substituting the values,
\[ M_x = \int_{0}^{1} \dfrac{1}{2} \{ x^4 – x^{1/4} \} \,dx \]
Solving the integration,
\[ M_x = \dfrac{1}{2} \Big{[} – \dfrac{x^5}{5} – \dfrac{4x^{5/4}}{5} \Big{]}_{0}^{1} \]
Substituting the limits,
\[ M_x = \dfrac{1}{2} \bigg{[} \Big{[} – \dfrac{1^5}{5} – \dfrac{4(1)^{5/4}}{5} \Big{]} – \Big{[} – \dfrac{0^5}{5} – \dfrac{4(0)^{5/4}}{5} \Big{]} \bigg{]} \]
Simplifying,\[ M_x = -0.3 \]
$y$-moment is given by,
\[ M_y = \int_{a}^{b} x \{ f(x) – g(x) \} \,dx \]
Substituting the values,
\[ M_y = \int_{0}^{1} x (x^4 – x^{1/4}) \,dx \]
\[ M_y = \int_{0}^{1} x^5 – x^{5/4} \,dx \]
Solving the integration,
\[ M_y = \Big{[} \dfrac{x^6}{6} – \dfrac{4x^{9/4}}{9} \Big{]}_{0}^{1} \]
\[ M_y = \Big{[} \dfrac{1^6}{6} – \dfrac{4(1)^{9/4}}{9} \Big{]} – \Big{[} \dfrac{0^6}{6} – \dfrac{4(0)^{9/4}}{9} \Big{]} \]
Simplifying,
\[ M_y = -0.278 \]
Now we can calculate the coordinates of the centroid $ ( \overline{x} , \overline{y} )$ using the above calculated values of Area and Moments of the region.
\[ \overline{x} = \dfrac{M_y}{A} \]
\[ \overline{x} = \dfrac{-0.278}{-0.6} \]
\[ \overline{x} = 0.463 \]
And,
\[ \overline{y} = \dfrac{M_x}{A} \]
\[ \overline{y} = \dfrac{-0.3}{-0.6} \]
\[ \overline{y} = 0.5 \]
Centroid of the Region $( \overline{x} , \overline{y} ) = (0.463, 0.5)$, which exactly points the center of the region in Figure 2.
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