Problem 21 - Single Variable Calculus

Question

Find the centroid of the region bounded by the graphs of the given equations. $$ y=x^{3}, \quad y=\sqrt[3]{x}, \quad x=0, \quad x=1 $$

Step-by-Step Solution

Answer

The centroid of the region bounded by the given equations is \(\left(\frac{61}{56}, \frac{16}{35}\right)\).

Step 1: Identify bounding curves and set up integrals

First, let's identify the curves bounding the region: - Left boundary: \(x=0\) - Right boundary: \(x=1\) - Lower curve: \(y=x^3\) - Upper curve: \(y=\sqrt[3]{x}\) Now we will set up the integrals for the x-coordinate of the centroid, \(\bar{x}\), and the y-coordinate of the centroid, \(\bar{y}\): \[- A=\int_a^b (f(x) - g(x)) dx \] \[ \bar{x} =\frac{1}{A}\int_a^b x(f(x) - g(x)) dx \] \[ \bar{y} =\frac{1}{A}\int_a^b \frac{1}{2}(f^2(x) - g^2(x)) dx \] Where: - \(f(x)\) = upper curve, in this case, \(\sqrt[3]{x}\) - \(g(x)\) = lower curve, in this case, \(x^3\) - \(a\) = left boundary, in this case, \(0\) - \(b\) = right boundary, in this case, \(1\)

Step 2: Find the area A

Now let's calculate the area A using the formula: \[ A = \int_0^1 (\sqrt[3]{x} - x^3) dx \] We can compute the integral: \[A = \left[\frac{3}{4}x^{4/3} - \frac{1}{4}x^4 \right]_0^1 =\left(\frac{3}{4}(1)^{4/3} - \frac{1}{4}(1)^4\right)-\left(\frac{3}{4}(0)-\frac{1}{4}(0)\right)=\frac{3}{4}-\frac{1}{4}=\frac{1}{2}\]

Step 3: Find the x-coordinate of the centroid, \(\bar{x}\)

Next, find the x-coordinate (\(\bar{x}\)) by calculating the integral: \[ \bar{x} = \frac{1}{A}\int_0^1 x(\sqrt[3]{x} - x^3) dx = \frac{2}{1}\int_0^1 x(\sqrt[3]{x} - x^3) dx = 2\int_0^1 x(\sqrt[3]{x} - x^3) dx \] We calculate the integral: \[ 2\int_0^1 x(\sqrt[3]{x} - x^3) dx = 2\left[ \frac{3}{7}x^{7/3} - \frac{1}{8}x^8\right]_0^1 = 2\left(\frac{3}{7}(1)^{7/3} - \frac{1}{8}(1)^8 - (0)\right) =2\left(\frac{3}{7} - \frac{1}{8}\right) = \frac{61}{56} \]

Step 4: Find the y-coordinate of the centroid, \(\bar{y}\)

Finally, find the y-coordinate (\(\bar{y}\)) of the centroid: \[ \bar{y} = \frac{1}{A}\int_0^1 \frac{1}{2}(\sqrt[3]{x^2} - x^6) dx = 2\int_0^1\frac{1}{2}(\sqrt[3]{x^2} - x^6) dx = \int_0^1 (\sqrt[3]{x^2} - x^6) dx \] We calculate the integral: \[ \int_0^1 (\sqrt[3]{x^2} - x^6) dx = \left[ \frac{3}{5}x^{5/3} - \frac{1}{7}x^7 \right]_0^1 = \left(\frac{3}{5}(1)^{5/3} - \frac{1}{7}(1)^7 \right) - \left(\frac{3}{5}(0)^{5/3} - \frac{1}{7}(0)^7 \right) = \frac{16}{35} \]

Step 5: Find the centroid's coordinates

Now that we have the x-coordinate \(\bar{x}\) and the y-coordinate \(\bar{y}\) of the centroid, the centroid is given by the coordinates: \[(\bar{x}, \bar{y}) = \left(\frac{61}{56}, \frac{16}{35}\right) \] So the centroid of the region bounded by the graphs of the given equations is \(\left(\frac{61}{56}, \frac{16}{35}\right)\).