the volume of the solid obtained by rotating the region enclosed by x=0, y=1, x=y^6
about the line y=1 can be computed using the method of disks or washers via an integral. find integral

We are asked to find the volume of the solid obtained by rotating the region enclosed by \(x=0\), \(y=1\), and \(x = y^6\) about the line \(y = 1\). We will use the method of disks or washers to set up the integral.

### Step 1: Understanding the region

- The boundaries are:
  - \(x = 0\) (the \(y\)-axis),
  - \(y = 1\) (a horizontal line),
  - \(x = y^6\) (a curve that describes the relationship between \(x\) and \(y\)).

The region is enclosed between \(x = 0\) and \(x = y^6\) and between \(y = 0\) and \(y = 1\).

### Step 2: Applying the washer method

Since we are rotating the region about the line \(y = 1\), the washer method is appropriate. The volume element will be the difference between the outer radius and the inner radius squared, integrated along \(y\).

The outer radius is the distance from the axis of rotation \(y = 1\) to the \(y = 0\) line, which is constant and equal to 1.

The inner radius is the distance from the axis of rotation \(y = 1\) to the curve \(x = y^6\). Since the rotation axis is horizontal, the distance from any point \((x, y)\) to the line \(y = 1\) is \(1 - y\).

The volume of the solid is given by the following integral:

\[
V = \pi \int_0^1 \left( 1^2 - (1 - y)^2 \right) \, dx
\]

#### Simplifying the integrand:

The expression inside the integral is:

\[
1^2 - (1 - y)^2 = 1 - (1 - 21

The volume of the solid obtained by rotating the region enclosed by x=0, y=1, x=y^6 about the line y=1 can be computed using the method of disks or washers via an integral. Find the integral.

This page covers the volume of a solid obtained by rotating the region enclosed by x=0, y=1, and x=y^6 about the line y=1. The solution is calculated using the washer method and integral calculus, providing a step-by-step explanation of the process. Ideal for students in undergraduate calculus.

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