Math Problem Statement

Using the disk method, determine the volume of a solid formed by revolving the region bounded above by the line y=1, on the left by the line x=0, and below by the curve the x=y^3/2 about the y=1 line . using disc method Setup and solve the integral that represents the volume of the solid of revolution described above

Solution

To solve this problem using the disk method, we need to set up the integral for the volume of the solid formed by revolving the region about the line y=1y = 1.

1. Identify the boundaries and curve:

  • The region is bounded:
    • Above by the line y=1y = 1
    • On the left by the line x=0x = 0
    • Below by the curve x=y3/2x = y^{3/2}

2. Transformation for the disk method:

Since we are revolving around the horizontal line y=1y = 1, the radius of each disk will be the distance between the curve and the line y=1y = 1.

The radius of a disk at a particular yy-coordinate is the vertical distance between the line y=1y = 1 and the curve x=y3/2x = y^{3/2}. Since we are using horizontal disks, the radius is given by:

r(y)=1yr(y) = 1 - y

The cross-sectional area of each disk will be:

A(y)=π[r(y)]2=π(1y)2A(y) = \pi [r(y)]^2 = \pi (1 - y)^2

We are revolving over the interval of yy, which is from y=0y = 0 to y=1y = 1.

3. Set up the volume integral:

The formula for the volume using the disk method is:

V=abπ[r(y)]2dyV = \int_a^b \pi [r(y)]^2 \, dy

For our problem, this becomes:

V=01π(1y)2dyV = \int_0^1 \pi (1 - y)^2 \, dy

4. Solve the integral:

First, expand the integrand:

(1y)2=12y+y2(1 - y)^2 = 1 - 2y + y^2

Thus, the volume integral becomes:

V=π01(12y+y2)dyV = \pi \int_0^1 (1 - 2y + y^2) \, dy

Now, integrate term by term:

V=π[011dy201ydy+01y2dy]V = \pi \left[ \int_0^1 1 \, dy - 2 \int_0^1 y \, dy + \int_0^1 y^2 \, dy \right]

Calculate each term:

011dy=y01=1\int_0^1 1 \, dy = y \Big|_0^1 = 1 01ydy=y2201=12\int_0^1 y \, dy = \frac{y^2}{2} \Big|_0^1 = \frac{1}{2} 01y2dy=y3301=13\int_0^1 y^2 \, dy = \frac{y^3}{3} \Big|_0^1 = \frac{1}{3}

Thus, the integral becomes:

[ V = \pi \left[ 1 - 2 \times \frac{1}{2} + \frac{1}{3} \right] = \pi \left[ 1 - 1 + \frac{1}{3} \right] = \pi \times \frac{1}{3} ]

5. Final answer:

The volume of the solid is:

V=π3V = \frac{\pi}{3}


Would you like further details or have any questions about this solution?

Related questions:

  1. How would the integral change if the solid were rotated around the x-axis instead?
  2. What if the region were rotated around the line y=2y = 2 instead of y=1y = 1?
  3. Can you apply the shell method to this problem for comparison?
  4. What is the effect of changing the curve to x=y2x = y^2 on the volume?
  5. How do we handle the volume if the region had different bounds on xx?

Tip:

When using the disk method, always ensure the radius is measured as the distance from the axis of rotation to the curve.

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Math Problem Analysis

Mathematical Concepts

Calculus
Volume of Solids of Revolution
Disk Method
Integral Calculus

Formulas

V = π ∫[a, b] [r(y)]² dy
r(y) = 1 - y

Theorems

Disk Method for Volume Calculation

Suitable Grade Level

Grades 11-12, College Level Calculus