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.
Figure-1 : Area of Centroid Curve
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.
Figure-2 : Equations of Moments
$x$-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,
\[ M_x = \dfrac{1}{2} \Big{[} \int_{0}^{1}Â x^6 \,dx – \int_{0}^{1} x^{2/3} \,dx \]
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 3.
Figure-3 : Centroid Region
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,
\[ M_x = \dfrac{1}{2} \bigg{[}Â \Big{[} – \dfrac{1^5}{5} – \dfrac{4(1)^{5/4}}{5} \Big{]} –Â \Big{[} – \dfrac{0^5}{5} – \dfrac{4(0)^{5/4}}{5} \Big{]} \bigg{]} \]
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..
Figure-4 : Centroid of Region
Images/Mathematical drawings are created with Geogebra.