Find the volume of the solid with given base and cross sections. The base is the unit circle x^2+y^2=1 and the cross sections perpendicular to the x-axis are triangles whose height and base are equal.

The base of a solid is bounded by x2 + y2 = 16. Cross sections perpendicular to the x-axis are right isosceles triangles with one leg located in the base. Which definite integral represents the volume of this solid?

To find the volume of the solid with the given properties, we can use integral calculus.

First, let's consider a cross-section of the solid. The cross-section is a triangle whose height and base are equal. Since the base is the unit circle, the height of the triangle is equal to the radius of the circle, which is 1.

Let's denote the variable along the x-axis as x, and let's consider a small segment of the x-axis with width Δx. The corresponding cross-section of the solid will have width Δx as well.

The area of the cross-section can be found by multiplying the base (Δx) by the height (1). So, the area of the cross-section is Δx.

To find the volume, we need to sum up the areas of all the cross-sections as Δx approaches 0 over the interval [-1, 1].

Using the integral notation, the volume (V) can be found as follows:

V = ∫[-1,1] Δx

Integrating Δx with respect to x over the interval [-1, 1] gives:

V = ∫[-1,1] 1 dx

Integrating 1 with respect to x gives:

V = x |[-1,1]

Evaluating the integral limits:

V = 1 - (-1)

V = 1 + 1

V = 2

Therefore, the volume of the solid with the given base and cross sections is 2 cubic units.

To find the volume of the solid, we can use the method of integral calculus.

First, let's visualize the solid. The base is the unit circle, and the cross sections perpendicular to the x-axis are triangles whose height and base are equal. This means that each triangle is an isosceles triangle.

To find the volume, we need to integrate the cross-sectional area function over the range of x-values that spans the base of the solid.

To begin, let's find the equation of one of the isosceles triangles. Since the height and base are equal, we can draw the altitude from the top vertex of the triangle (which lies on the x-axis) to the base of the triangle. This altitude will split the base into two equal line segments (each of length b/2, where b is the length of the base).

Now, let's consider a triangle at a specific x-value. We know that the height and base are equal, and the altitude from the top vertex to the base divides the base into two segments of length b/2.

The equation of the line formed by the altitude passing through the top vertex is x = x-coordinate of the top vertex. Since the top vertex lies on the x-axis, its x-coordinate is zero. Therefore, the equation of the line is x = 0.

The length of the base of the triangle is twice the x-coordinate of the point where the line intersects the circle (since the triangle is symmetric about the y-axis). Solving for the points of intersection between the line x = 0 and the circle x^2 + y^2 = 1, we find two points: (0, 1) and (0, -1).

Thus, the length of the base of the triangle is 2(-1 - 0) = 2. Since the height and base of the triangle are equal, the area of the triangle is (1/2) * base * height = (1/2) * 2 * 2 = 2.

Now that we have the equation for the area of the triangle at a specific x-value, we can express it as a function of x and integrate it over the range of x-values that spans the base of the solid (from -1 to 1, since the circle is centered at the origin with radius 1).

Therefore, the volume of the solid is given by the integral:

V = ∫[from -1 to 1] 2 dx

Evaluating the integral, we get:

V = [2x] [from -1 to 1]
= 2(1) - 2(-1)
= 2 + 2
= 4

Hence, the volume of the solid is 4 cubic units.