A volume is described as follows:
The base is the region bounded by  
y
=
5
−
5
x
2
  and  
y
=
0
,
  and every cross section parallel to the  
x
 -axis is a right triangle whose height and base are equal, and the base of the right triangle lies along the base of the region.

Find the volume of this object.

Volume =

To find the volume of the given solid, we need to use the method of cross-sections. Let's break it down step by step:

### Step 1: Understand the Region
The base of the solid is bounded by the curve \( y = 5 - 5x^2 \) and the line \( y = 0 \), which is the x-axis.

This region is a parabola opening downward with its vertex at \( (0, 5) \) and crossing the x-axis at \( x = \pm 1 \) (since \( 5 - 5x^2 = 0 \) gives \( x = \pm 1 \)).

### Step 2: Cross-sectional Area
The cross sections of the solid are described as right triangles, with the base of the triangle lying along the region defined by the curve and the x-axis. Importantly, the height and the base of each right triangle are equal.

At any point \( x \), the base of the right triangle is the distance from the curve to the x-axis, which is \( y = 5 - 5x^2 \).

Thus, the base of the triangle is:
\[
b(x) = 5 - 5x^2
\]
Since the height equals the base, the height is also:
\[
h(x) = 5 - 5x^2
\]

The area \( A(x) \) of the triangle is given by the formula for the area of a triangle:
\[
A(x) = \frac{1}{2} \times \text{base} \times \text{height}
\]
Substituting \( b(x) \) and \( h(x) \) into this:
\[
A(x) = \frac{1}{2} \times (5 - 5x^2) \times (5 - 5x^2) = \frac{1}{2} \times (5 - 5x^2)^2
\]

### Step 3: Set up the Integral
To find the volume, we integrate the cross-sectional area along the x-axis from \( x = -1 \) to \( x = 1 \):
\[
V = \int_{-1}^{1} A(x) \, dx = \int_{-1}^{1} \frac{1}{2} (5 - 5x^2)^2 \, dx
\]

### Step 4: Solve the Integral
First, expand \( (5 - 5x^2)^2 \):
\[
(5 - 5x^2)^2 = 25 - 50x^2 + 25x^4
\]

Now, substitute this into the integral:
\[
V = \frac{1}{2} \int_{-1}^{1} (25 - 50x^2 + 25x^4) \, dx
\]

Since the integrand is an even function, we can simplify the limits and multiply by 2:
\[
V = \int_{0}^{1} (25 - 50x^2 + 25x^4) \, dx
\]

Now integrate term by term:
\[
V = \left[ 25x - \frac{50x^3}{3} + \frac{25x^5}{5} \right]_0^1
\]

Evaluating at \( x = 1 \) and \( x = 0 \):
\[
V = \left( 25(1) - \frac{50(1)^3}{3} + \frac{25(1)^5}{5} \right) - \left( 25(0) - \frac{50(0)^3}{3} + \frac{25(0)^5}{5} \right)
\]
\[
V = 25 - \frac{50}{3} + 5 = 30 - \frac{50}{3}
\]

Simplifying the result:
\[
V = \frac{90}{3} - \frac{50}{3} = \frac{40}{3}
\]

### Final Answer:
The volume of the solid is:
\[
\boxed{\frac{40}{3}}
\]

Would you like to explore the method of cross-sections further or ask any questions?

### Related Questions:
1. How is the method of cross-sections applied to solids with different cross-sectional shapes?
2. What is the difference between using disks, washers, and cross-sections to find volume?
3. Can we apply this technique to solids with circular cross-sections?
4. How does the symmetry of the region affect the computation of the volume?
5. How do you find the volume of a solid when the cross sections are equilateral triangles?

### Tip:
Always check the symmetry of the region to potentially simplify your volume calculations by adjusting the limits of integration.

A volume is described as follows: The base is the region bounded by y = 5 - 5x^2 and y = 0, and every cross section parallel to the x-axis is a right triangle whose height and base are equal, and the base of the right triangle lies along the base of the region. Find the volume of this object.

This detailed solution explains how to find the volume of a solid whose base is bounded by y = 5 - 5x^2 and y = 0. The cross sections are right triangles with equal height and base. Learn how to use integral calculus and the method of cross-sections to solve this problem step by step. Suitable for advanced calculus students in Grades 11-12.

Math Problem Statement

A volume is described as follows: The base is the region bounded by
y

5 − 5 x 2 and
y

Solution

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