Question

Find the volume formed by rotating about the y-axis the region enclosed by: x=7y and y^3=x with y≥0

Answer 1

To calculate the volume formed by rotating the region enclosed by x=7y and y^3=x around the y-axis, one should use the washer method to integrate the area of a washer between the curves from the lower to the upper bound of y.

Step-by-step explanation:

Calculating the Volume of Revolution Around the Y-Axis

To find the volume formed by rotating the region enclosed by x=7y and y^3=x around the y-axis with y≥0, one could use the method of cylindrical shells. However, since this is a rotation around the y-axis, and we have a constraint y≥0, the washer method (or disc method with subtraction for the hole in the middle) becomes more appropriate. This involves integrating the area of a representative washer between the two curves from the lower to the upper bound of y.

We find the bounds of integration by setting the two functions equal to each other and solving for y: 7y=y^3. After finding the bounds, we can integrate the volume using the formula for the volume of revolution, V=∫π(R^2-r^2)dy, where R is the outer radius and r is the inner radius, corresponding to the functions given. This application of calculus allows us to find the volume of the shape formed by the rotation.

The complete question is:content loaded

Find the volume formed by rotating about the y-axis the region enclosed by: x=7y and y^3=x with y≥0 is:

Answer 2

The volume of the region between these two curves, which is the volume we're interested in, is the difference between these two volumes, which is approximately
`\( 543.04 \)` cubic units.

The volume of the solid formed by rotating a region around the y-axis can be found using the disk or washer method. The method you use depends on the shape of the region being rotated. In this case, we have two functions,
`\( x = 7y \)` and
`\( y^3 = x \)`, and we need to find their intersection points and the limits of integration.

First, we'll set the two equations equal to each other to find the points of intersection:

`\[ 7y = y^3 \]`

Then we will solve for \( y \). After finding the points of intersection, we can integrate with respect to \( y \) to find the volume. The volume \( V \) of the solid formed by rotating around the y-axis is given by:

`\[ V = \pi \int_(a)^(b) [f(y)]^2 dy \]`

where
`\( [f(y)]^2 \)` is the square of the function that gives the radius of the disk at a particular
`\( y \)`, and
`\( a \)` and
`\( b \)` are the limits of integration determined by the intersection points.

Let's start by finding the intersection points.

The intersection points between
`\( x = 7y \)` and
`\( y^3 = x \)` are
`\( y = 0 \)` and
`\( y \approx 2.646 \)` (rounded to three decimal places).

Using these points as the limits of integration, we'll calculate the volume formed by rotating the region enclosed by these two functions around the y-axis using the disk method.

However, it seems that the calculation has led to a result of zero, which is not expected. This suggests there might be an error in the calculation process. Let's re-evaluate the integral for the volume.

The volume of the solid of revolution, when rotating \( x = 7y \) around the y-axis from
`\( y = 0 \)` to
`\( y = 2.646 \)`, is:

`\[ V = \pi \int_(0)^(2.646) (7y)^2 dy \]`

And the volume of the solid when rotating
`\( y^3 = x \)` around the y-axis from
`\( y = 0 \)` to
`\( y = 2.646 \)` is:

`\[ V = \pi \int_(0)^(2.646) (y^3)^2 dy \]`

We need to subtract the second volume from the first to find the volume of the region between the curves. Let's do this calculation correctly.

The correct volumes are as follows:

The volume formed by rotating
`\( x = 7y \)` around the y-axis from
`\( y = 0 \)` to
`\( y \approx 2.646 \)` is approximately
`\( 950.32 \)` cubic units.

The volume formed by rotating
`\( y^3 = x \)` around the y-axis from
`\( y = 0 \)` to
`\( y \approx 2.646 \)` is approximately
`\( 407.28 \)` cubic units.

The volume of the region between these two curves, which is the volume we're interested in, is the difference between these two volumes, which is approximately
`\( 543.04 \)` cubic units.

This volume represents the solid formed by rotating the region enclosed by
`\( x = 7y \)` and
`\( y^3 = x \)` around the y-axis, with
`\( y \geq 0 \)`.