We are tasked with finding the volume $V$ of a solid whose base is an elliptical region, and cross-sections perpendicular to the $x$-axis are isosceles right triangles. Let's break this down step by step.

Step 1: Equation of the Ellipse

The boundary of the base of the solid is given by: $16x^2 + 9y^2 = 144$ Dividing through by 144, we get: $\frac{x^2}{9} + \frac{y^2}{16} = 1$ This is the equation of an ellipse with semi-major axis $4$ along the $y$-axis and semi-minor axis $3$ along the $x$-axis.

Step 2: Cross-Section Properties

The cross-sections perpendicular to the $x$-axis are isosceles right triangles with the hypotenuse lying along the base (i.e., the ellipse). In an isosceles right triangle, the relationship between the hypotenuse $h$ and the legs $s$ is: $s = \frac{h}{\sqrt{2}}$ Thus, for each cross-section, the area $A$ of the triangle is: $A = \frac{1}{2} s^2 = \frac{1}{2} \left( \frac{h}{\sqrt{2}} \right)^2 = \frac{h^2}{4}$

Step 3: Expression for $h$ in Terms of $x$

The hypotenuse of the triangle in each cross-section is the distance between the points on the ellipse for a fixed $x$. Solving the equation of the ellipse for $y$ gives: $y = \pm \frac{4}{3} \sqrt{9 - x^2}$ The total length of the hypotenuse is then: $h(x) = 2 \left( \frac{4}{3} \sqrt{9 - x^2} \right) = \frac{8}{3} \sqrt{9 - x^2}$

Step 4: Area of Cross-Sections

The area of each cross-section at a given $x$ is: $A(x) = \frac{h(x)^2}{4} = \frac{1}{4} \left( \frac{8}{3} \sqrt{9 - x^2} \right)^2 = \frac{64}{36} (9 - x^2) = \frac{16}{9} (9 - x^2)$

Step 5: Volume of the Solid

The volume $V$ of the solid is the integral of the areas of the cross-sections from $x = -3$ to $x = 3$ (the semi-minor axis bounds of the ellipse): $V = \int_{-3}^{3} A(x) \, dx = \int_{-3}^{3} \frac{16}{9} (9 - x^2) \, dx$

Step 6: Solving the Integral

We can break this integral into two simpler integrals: $V = \frac{16}{9} \left( \int_{-3}^{3} 9 \, dx - \int_{-3}^{3} x^2 \, dx \right)$

1. The first integral is straightforward: $\int_{-3}^{3} 9 \, dx = 9 \times (3 - (-3)) = 9 \times 6 = 54$
2. The second integral can be computed using the standard formula for $\int x^2 \, dx$ over symmetric limits: $\int_{-3}^{3} x^2 \, dx = 2 \int_0^3 x^2 \, dx = 2 \left[ \frac{x^3}{3} \right]_0^3 = 2 \times \frac{27}{3} = 18$

Now, substitute these values back into the expression for $V$: $V = \frac{16}{9} \left( 54 - 18 \right) = \frac{16}{9} \times 36 = 64$

Final Answer:

The volume of the solid is $V = 64$ cubic units.

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Do you want further details or have any questions? Here are five related questions:

1. How is the volume formula for solids with triangular cross-sections derived?
2. What happens if the cross-sections were equilateral triangles instead?
3. How do we compute the volume of a solid with a circular base instead of an ellipse?
4. How does the relationship between the base shape and cross-section type affect the volume?
5. What are other applications of solids with known cross-sectional areas?

Tip: Symmetry in geometric shapes often simplifies integral calculations, as seen with even functions over symmetric intervals.