Math Problem Statement

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.

Solution

The problem asks to find the volume of a solid whose base is the region bounded by the curves y=x2y = x^2 and y=xy = \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=x2y = x^2 (a parabola) and y=xy = \sqrt{x} (a square root function). The region of interest is where these two curves intersect.

    To find the intersection points: x2=xx^2 = \sqrt{x} Squaring both sides to eliminate the square root: x4=xx^4 = x Rearranging this equation: x4x=0x^4 - x = 0 Factor the equation: x(x31)=0x(x^3 - 1) = 0 This gives x=0x = 0 or x3=1x^3 = 1, so x=1x = 1.

    Thus, the region is bounded between x=0x = 0 and x=1x = 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 xx, which is: Side length=xx2\text{Side length} = \sqrt{x} - x^2 Since the area of a square is side length2\text{side length}^2, the area of the cross-section at xx is: A(x)=(xx2)2A(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=0x = 0 to x=1x = 1: V=01(xx2)2dxV = \int_0^1 \left( \sqrt{x} - x^2 \right)^2 dx

  4. Simplify the Integrand: Expand (xx2)2\left( \sqrt{x} - x^2 \right)^2: (xx2)2=x2x5/2+x4\left( \sqrt{x} - x^2 \right)^2 = x - 2x^{5/2} + x^4

  5. Integrate: Now, integrate each term separately: V=01(x2x5/2+x4)dxV = \int_0^1 \left( x - 2x^{5/2} + x^4 \right) dx

    • 01xdx=12\int_0^1 x \, dx = \frac{1}{2}
    • 01x5/2dx=27\int_0^1 x^{5/2} \, dx = \frac{2}{7}
    • 01x4dx=15\int_0^1 x^4 \, dx = \frac{1}{5}

    Therefore: V=122×27+15V = \frac{1}{2} - 2 \times \frac{2}{7} + \frac{1}{5} Simplify: V=1247+15V = \frac{1}{2} - \frac{4}{7} + \frac{1}{5} Finding a common denominator (70): V=35704070+1470V = \frac{35}{70} - \frac{40}{70} + \frac{14}{70} V=3540+1470=970V = \frac{35 - 40 + 14}{70} = \frac{9}{70}

Thus, the volume of the solid is 970\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.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Integration
Area Between Curves
Volumes of Solids

Formulas

Volume = ∫(Area of Cross Section) dx
Area of Cross Section = (Side Length)^2
Side Length = √x - x^2

Theorems

Fundamental Theorem of Calculus

Suitable Grade Level

Grade 11-12 (High School Calculus) or Early College Calculus