Find the volume V of the described solid S.
The base of a solid S is an elliptical region with boundary curve 16x^2 + 9y^2 = 144. Cross-sections perpendicular to the x-axis are isosceles right triangles with hypotenuse in the base

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.

---

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.

Find the volume V of the described solid S. The base of a solid S is an elliptical region with boundary curve 16x^2 + 9y^2 = 144. Cross-sections perpendicular to the x-axis are isosceles right triangles with hypotenuse in the base.

Explore how to calculate the volume of a solid with an elliptical base and isosceles right triangle cross-sections perpendicular to the x-axis. This step-by-step solution involves integral calculus and geometry, making it suitable for college-level students. The problem includes key formulas such as the equation of an ellipse and triangle area, leading to the final volume V = 64 cubic units.

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