\section*{Problem 11 - V01211}

The left arc of an ellipse (center at the origin) and its tangent at \(P_{1}(2 ; 3)\) bound a region with the x-axis, which, when rotated about the x-axis, forms a drop-shaped body with the largest cross-sectional area \(q=12 \pi\) square units.

What is the volume of the solid of revolution?
}

In the figure, the area rotating around the x-axis is highlighted in color. Its largest cross-section is a circle and occurs at \(x=0\), from which \(\pi b^{2}=12 \pi\) immediately gives \(b=\sqrt{12}\).

From the ellipse equation \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\), with \(b^{2}=12\), substituting the point \(P(2 ; 3)\) yields that the major semi-axis \(a=4\), i.e., \(a^{2}=16\). Using the tangent equation \(\frac{x x_{P}}{a^{2}}+\frac{y y_{P}}{b^{2}}=1\), the tangent becomes

\[
\frac{2 x}{16}+\frac{3 y}{12}=1 \rightarrow y=4-\frac{x}{2}
\]

with the zero at \(x=8\).

![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-1711.jpg?height=505&width=930&top_left_y=2009&top_left_x=563)

Thus, the rotating area consists of a right triangle (cathetus lengths \(3 \mathrm{~cm}\) and 6 \(\mathrm{cm}\)) and an elliptical segment from \(x=-4\) to \(x=2\) of the ellipse \(\frac{x^{2}}{16}+\frac{y^{2}}{12}=1\). It follows that

\[
\begin{aligned}
V & =V_{\text {Triangle }}+V_{\text {Segment }}=\pi \cdot 3^{2} \cdot 6+\int_{-4}^{2}\left[\frac{1}{4} \sqrt{192-12 x^{2}}\right]^{2} d x \\
& =54 \pi+\left[\frac{x}{4}\left(48-x^{2}\right)\right]_{-4}^{2}=54 \pi+54
\end{aligned}
\]

The solid of revolution has a volume of \(54(\pi+1) \approx 223.6 \mathrm{~cm}^{3}\).