Problem 79 - Thomas Calculus

Question

Centroid Find the centroid of the region that is bounded below by the \(x\) -axis and above by the ellipse \(\left(x^{2} / 9\right)+\left(y^{2} / 16\right)=1\)

Step-by-Step Solution

Answer

The centroid is \( \left(\frac{32}{9}, \frac{128}{15\pi}\right) \).

Step 1: Determine the Limits of Integration

The given ellipse equation is \( \frac{x^2}{9} + \frac{y^2}{16} = 1 \). Observe that the ellipse is symmetric about both axes, implying we only need to find the centroid of one quadrant and multiply appropriately. The limits for \(x\) are from \(-3\) to \(3\) and for \(y\), from \(0\) to \(4\).

Step 2: Identify the Area of the Region

First, calculate the area of the region using the formula for ellipses, \( \pi ab \), where \(a = 3\) and \(b = 4\). Thus, the area is \( 12\pi \). Then, to find the area in the first quadrant, we take one-fourth of this area: \( \frac{12\pi}{4} = 3\pi \).

Step 3: Set Up the Integral for \(x\)-Coordinate of the Centroid

The \(x\)-coordinate of the centroid, \(\bar{x}\), is given by \( \frac{1}{A} \int_{-a}^{a} x \int_0^{b}\! dydx \). Since only the first quadrant is used, simplify to \( \frac{4}{A} \int_0^{3} x \left(\int_0^{\sqrt{16(1 - \frac{x^2}{9})}} dy \right) dx \), resulting in \( \frac{4}{A} \int_0^{3} x (2b \sqrt{1-\frac{x^2}{a^2}}) dx \).

Step 4: Calculate the \(x\)-Coordinate of the Centroid

Calculate \( \bar{x} = \frac{4}{3\pi} \int_0^3 x \sqrt{16 - \frac{16}{9}x^2} dx \). Simplifying, the integral becomes \( \frac{64}{9\pi} \int_0^3 x \sqrt{9 - x^2} dx = \frac{64}{9\pi} \left[-\frac{x}{2} \sqrt{9-x^2} - \frac{9}{2} \arcsin{(\frac{x}{3})} \right]_0^3 \rightarrow \frac{32}{9}\).

Step 5: Repeat the Process for \(y\)-Coordinate

Similarly, for the \(y\)-coordinate of the centroid, \(\bar{y}\), use \( \bar{y} = \frac{4}{A} \int_0^3 \left(\frac{1}{2 \sqrt{1 - \frac{x^2}{a^2}}} \int_0^{\sqrt{16(1 - \frac{x^2}{9})}} y^2 dy \right) dx \), simplifying to \( \frac{128}{15\pi} \).

Step 6: State the Final Result for the Centroid

Combining results from both coordinates, the centroid of the region is at \( \left(\frac{32}{9}, \frac{128}{15\pi}\right) \).

Key Concepts

limits of integration

When finding the centroid of a region bounded by an ellipse, determining the limits of integration is crucial. In our case, the given ellipse is expressed as \(\frac{x^2}{9} + \frac{y^2}{16} = 1\). This specific equation hints at an ellipse that stretches along both the x and y axes with semi-axes lengths of 3 and 4, respectively. The symmetry of the ellipse around both axes allows us to focus on just one quadrant for simplicity. The x-coordinate limits are determined based on the maximum and minimum values that x can take within the ellipse, which are from -3 to 3. Meanwhile, the y-coordinate values range from 0 to 4, stemming from the equation structure when x equals zero. The symmetry of the ellipse reduces the calculus work, making the integration over one quadrant representative for the entire shape's centroid calculation. This is an effective strategy for integrating complex forms.

x-coordinate of centroid

To find the x-coordinate of the centroid \(\bar{x}\), the integration involves both geometric intuition and mathematical precision. The formula used is: \[\bar{x} = \frac{4}{A} \int_0^3 x \left(2b \sqrt{1-\frac{x^2}{a^2}}\right) dx\]Here, \(A\) represents the area of the ellipse within the first quadrant (\(3\pi\)), which simplifies the problem into focusing just on this one piece. The integral collects all x-moments within this region according to its horizontal distance. The expression inside the integral, \(x(2b \sqrt{1-\frac{x^2}{a^2}})\), accounts for the ellipse's vertical stretch and positional symmetry along the x-axis. Evaluating this integral through substitution or a straightforward application of known trigonometric identities helps derive the exact x-coordinate of the centroid at \(\frac{32}{9}\). This precise calculation gives us information about how the mass of the area is distributed horizontally.

y-coordinate of centroid

Calculating the y-coordinate of the centroid \(\bar{y}\) is similar in intent, but the integration involves y-components focused in the first quadrant of the ellipse. We employ the formula:\[\bar{y} = \frac{4}{A} \int_0^3 \left(\frac{1}{2 \sqrt{1 - \frac{x^2}{a^2}}} \int_0^{\sqrt{16(1 - \frac{x^2}{9})}} y^2 dy\right) dx\]This formula evaluates how the region's area relates vertically, by accumulating the vertical distances squared \(y^2\) and adjusting for the shape's distribution along the y-axis. The outer integral respects the symmetry along x by scaling the results with the ellipse's vertical constraints at each point, akin to employing layers of heights within the quadrant.Finally, upon executing these calculations, we find \(\bar{y}\) to be \(\frac{128}{15\pi}\), painting the full picture of where the centroid point is located within the y-axis in the ellipse’s interior, offering insights into the vertical center of the first quadrant's mass.