https://file.aiquickdraw.com/mathbot/file/1727021271801ngey0n71.jpg

The problem asks to find the volume of a solid whose base is the region bounded by the curves \( y = x^2 \) and \( y = \sqrt{x} \), with cross-sections perpendicular to the x-axis being squares.

### Steps to solve the problem:

1. **Understand the Region**:
   The curves given are \( y = x^2 \) (a parabola) and \( y = \sqrt{x} \) (a square root function). The region of interest is where these two curves intersect. 
   
   To find the intersection points:
   \[
   x^2 = \sqrt{x}
   \]
   Squaring both sides to eliminate the square root:
   \[
   x^4 = x
   \]
   Rearranging this equation:
   \[
   x^4 - x = 0
   \]
   Factor the equation:
   \[
   x(x^3 - 1) = 0
   \]
   This gives \( x = 0 \) or \( x^3 = 1 \), so \( x = 1 \).

   Thus, the region is bounded between \( x = 0 \) and \( x = 1 \).

2. **Cross-Sectional Area**:
   The cross-sections perpendicular to the x-axis are squares. The side length of each square is the vertical distance between the two curves at a given \( x \), which is:
   \[
   \text{Side length} = \sqrt{x} - x^2
   \]
   Since the area of a square is \( \text{side length}^2 \), the area of the cross-section at \( x \) is:
   \[
   A(x) = \left( \sqrt{x} - x^2 \right)^2
   \]

3. **Volume Calculation**:
   To find the volume of the solid, integrate the area of the cross-sections from \( x = 0 \) to \( x = 1 \):
   \[
   V = \int_0^1 \left( \sqrt{x} - x^2 \right)^2 dx
   \]

4. **Simplify the Integrand**:
   Expand \( \left( \sqrt{x} - x^2 \right)^2 \):
   \[
   \left( \sqrt{x} - x^2 \right)^2 = x - 2x^{5/2} + x^4
   \]

5. **Integrate**:
   Now, integrate each term separately:
   \[
   V = \int_0^1 \left( x - 2x^{5/2} + x^4 \right) dx
   \]
   - \( \int_0^1 x \, dx = \frac{1}{2} \)
   - \( \int_0^1 x^{5/2} \, dx = \frac{2}{7} \)
   - \( \int_0^1 x^4 \, dx = \frac{1}{5} \)

   Therefore:
   \[
   V = \frac{1}{2} - 2 \times \frac{2}{7} + \frac{1}{5}
   \]
   Simplify:
   \[
   V = \frac{1}{2} - \frac{4}{7} + \frac{1}{5}
   \]
   Finding a common denominator (70):
   \[
   V = \frac{35}{70} - \frac{40}{70} + \frac{14}{70}
   \]
   \[
   V = \frac{35 - 40 + 14}{70} = \frac{9}{70}
   \]

Thus, the volume of the solid is \( \frac{9}{70} \).

Would you like further details or clarifications on any of these steps?

### Related Questions:
1. What is the general formula for calculating volumes using cross-sectional areas?
2. How do we handle cross-sections of different shapes, such as equilateral triangles or semicircles?
3. How does the region of intersection between curves affect the volume of a solid?
4. What happens to the volume if the cross-sections are parallel to the y-axis instead of the x-axis?
5. How would the calculation change if the cross-sections were rectangles instead of squares?

### Tip:
Always check the bounds of integration carefully, especially when curves intersect, to ensure you're integrating over the correct region.

Find the volume of the solid whose base is the region bounded by the curves y = x^2 and y = √x and with cross sections perpendicular to x-axis are squares.

Learn how to calculate the volume of a solid with square cross sections, with its base formed by the region between the curves y = x^2 and y = √x. This problem involves integrating the area between curves and calculating the volume of the resulting solid. Suitable for high school calculus or early college students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation