Questions: Find the volume of the solid with the given base and cross sections. The base is the circle x^2 + y^2 = 16, and the cross sections perpendicular to the x-axis are triangles whose height and base are equal.

Solution Steps

To find the volume of the solid, we need to integrate the area of the cross-sectional triangles along the x-axis. The base of the solid is a circle with equation \(x^2 + y^2 = 16\), which implies the radius is 4. The cross-sections perpendicular to the x-axis are isosceles right triangles with equal height and base. The length of the base of each triangle at a given x is twice the y-coordinate, \(2\sqrt{16 - x^2}\), and the area of each triangle is \(\frac{1}{2} \times \text{base} \times \text{height}\). Integrate this area from \(x = -4\) to \(x = 4\).

The base of the solid is a circle given by the equation \(x^2 + y^2 = 16\). This implies that the radius of the circle is 4.

The cross-sections perpendicular to the \(x\)-axis are isosceles right triangles. The base of each triangle at a given \(x\) is \(2\sqrt{16 - x^2}\), which is twice the \(y\)-coordinate. The area of each triangle is given by: \[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times \left(2\sqrt{16 - x^2}\right)^2 = 32 - 2x^2 \]

To find the volume of the solid, integrate the area of the cross-sections from \(x = -4\) to \(x = 4\): \[ \text{Volume} = \int_{-4}^{4} (32 - 2x^2) \, dx \]

Evaluating the integral, we find: \[ \text{Volume} = 170.6667 \]

Final Answer