Question

The volume that has a base of the ellipse \(x^{2} / 4+y^{2} / 9=1\) and cross- sections of an equilateral triangle perpendicular to the \(y\) -axis. Use the method of slicing.

Short answer

The volume of the solid is \( 24\sqrt{3} \).

Step-by-step solution

Identify the Ellipse Equation

The given equation of the ellipse is \( \frac{x^{2}}{4} + \frac{y^{2}}{9} = 1 \). This corresponds to the standard form \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) where \( a = 2 \) and \( b = 3 \). The major axis is along the \( y \)-axis.

Find the Semi-Major Axis Length

Since the major axis is along the \( y \)-axis, the semi-major axis length \( b = 3 \). Cross-sections perpendicular to the \( y \)-axis will vary with the ellipse's equation.

Solve for \( x \)

Rearrange the ellipse equation to express \( x \) in terms of \( y \). Solve: \( \frac{x^2}{4} = 1 - \frac{y^2}{9} \) and then \( x^2 = 4 \left( 1 - \frac{y^2}{9} \right) \).

Calculate Cross-Section Width

Each cross-section is an equilateral triangle. The width of the triangle at any \( y \) is the length \( 2x \). Substitute: \( x = \sqrt{4 - \frac{4y^2}{9}} \) giving width \( 2x = 2\sqrt{4 - \frac{4y^2}{9}} \).

Determine the Area of Cross-Section

The area of an equilateral triangle with side length \( s \) is \( \frac{\sqrt{3}}{4} s^{2} \). For the width \( 2x \), \( A(y) = \frac{\sqrt{3}}{4} (2\sqrt{4 - \frac{4y^2}{9}})^{2} \).

Compute the Volume

Integrate the area of the cross-sections along the \( y \)-axis from \(-b\) to \(b\). The bounds are from -3 to 3: \( V = \int_{-3}^{3} \frac{\sqrt{3}}{4} (2\sqrt{4 - \frac{4y^2}{9}})^{2} \ dy \).

Evaluate the Integral

Calculate the integral: \( V = \sqrt{3} \int_{-3}^{3} (4 - \frac{4y^2}{9}) \ dy \). Factor out constants and simplify to complete the integral, which results in multiplying by the obtained factors of \( y \) evaluated between the limits.

Key concepts

Ellipse

An ellipse is like a stretched circle, perfectly represented by the equation \(\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1\). This equation showcases how the ellipse elongates in the plane, with \(a\) and \(b\) indicating the semi-axes, or the distances from the center to the edge of the ellipse. For this problem, we find \(a = 2\) and \(b = 3\). The longest axis, which is called the major axis, in this case, runs parallel to the \(y\)-axis, making \(b\) the semi-major axis length. This geometry is essential when you set up the base for constructing solids.

Volume of a Solid

The volume of a solid can often be found using a slicing method, especially when regular cross-sections (like circles or triangles) run throughout the object. Imagine a loaf of bread, where each slice is the same shape as the previous one. To find the volume, you calculate the area of one slice — the cross-section — and "stack" them along the axis of interest. In this exercise, each slice perpendicular to the \(y\)-axis is an equilateral triangle. Therefore, understanding the shape and area of this triangle for every vertical slice of the ellipse is crucial in determining the total volume of the solid.

Integration

Integration is the mathematical tool we use to compute the sum of infinitely small quantities, like adding up all the slices to find total volume. In this task, the equation for the cross-sectional area, \(A(y)\), changes with \(y\). By integrating \(A(y)\) from \(-3\) to \(3\), we effectively sum up the areas of all triangular slices. This integration respects the ellipse's boundary, ensuring each slice's area respects its actual size determined by \(x\), where \(x = \sqrt{4 - \frac{4y^2}{9}}\). Thus, integration provides a full, accurate volume of the solid.

Cross-Section Area

Every cross-section here is an equilateral triangle whose sides change as \(y\) changes. The width of each triangle is linked to the ellipse, defaulting to \(2x\). For an equilateral triangle, knowing one side lets you determine the area using the formula \(\frac{\sqrt{3}}{4} s^{2}\). Plugging in \(s = 2x\), we get \(A(y) = \frac{\sqrt{3}}{4} (2\sqrt{4 - \frac{4y^2}{9}})^{2}\). Calculating this at each slice, and then integrating, connects each discrete slice into a continuous volume, neatly fitting them within the phase of the ellipse.