Question

Sketch the region \(R\) bounded by the graphs of the given equations and show a typical horizontal slice. Find the volume of the solid generated by revolving \(R\) about the \(y\) -axis. \(x=y^{3 / 2}, y=9, x=0\)

Short answer

Volume is \(3280.5\pi\).

Step-by-step solution

Sketch the Region

We need to sketch the region bounded by the given equations: \(x = y^{3/2}\), \(y = 9\), and \(x = 0\). - The curve \(x = y^{3/2}\) represents a sideways parabola opening to the right.- The line \(y = 9\) is a horizontal line, acting as an upper bound.- The line \(x = 0\) is the y-axis, acting as a left bound.The region \(R\) is the area between these curves from \(y = 0\) to \(y = 9\).

Set Up the Integral for Volume

The volume of the solid formed by revolving region \(R\) around the \(y\)-axis can be calculated using the method of cylindrical shells.The formula for the volume with cylindrical shells is:\[V = 2\pi \int_{a}^{b} (\text{radius})(\text{height}) \, dy\]In this scenario:- Radius = \(x = y^{3/2}\).- Height = \(x - 0 = y^{3/2}\).Thus, the volume integral becomes:\[V = 2\pi \int_{0}^{9} y^{3/2} \cdot y^{3/2} \, dy\]Simplify this to:\[V = 2\pi \int_{0}^{9} y^{3} \, dy\]

Compute the Integral

Now, integrate the function \(y^3\) from 0 to 9:\[V = 2\pi \int_{0}^{9} y^3 \, dy = 2\pi \left[ \frac{y^4}{4} \right]_{0}^{9}\]Calculate the definite integral:- At \(y = 9\), \(\frac{9^4}{4} = \frac{6561}{4}\).- At \(y = 0\), the expression evaluates to 0.Thus, the calculated integral evaluates to:\[V = 2\pi \left( \frac{6561}{4} - 0 \right) = 2\pi \cdot \frac{6561}{4}\]

Final Calculation

Compute the final volume:- \(V = 2\pi \cdot \frac{6561}{4} = \frac{13122\pi}{4} = 3280.5\pi\)Thus, the volume of the solid is \(3280.5\pi\).

Key concepts

Cylindrical Shells Method

The Cylindrical Shells Method is a useful technique for finding the volume of a solid of revolution. Imagine you have a region in two-dimensional space that you want to "spin around" a vertical line, creating a three-dimensional solid. Instead of slicing this solid into discs or washers—like in the disc method—the cylindrical shells method involves wrapping thin vertical bands around the axis of rotation to create thin, hollow cylinders.
* **Radius**: The radius of each cylindrical shell is the horizontal distance from the vertical axis of rotation to the curve. In our case, when rotating around the y-axis, this distance is represented by the x-coordinate of a point on the curve.* **Height**: This is the function value given by the equation of the curve. For the equation given, the height of the shell is also the y-coordinate.
The volume of each shell is calculated by the formula \[ V = 2\pi imes ( ext{radius}) imes ( ext{height}) \]where the integral sums up the volumes of all such infinitesimally thin shells as they vary along the axis of revolution. By setting up the integral, we essentially stack these tiny shell volumes together to find the complete volume of the solid.

Definite Integral

The definite integral is a mathematical tool used to calculate the total accumulation of quantities, such as areas and volumes, over a specified interval. It not only helps in finding areas under curves but also in computing volumes of solids, especially when the shapes are not standard or symmetrical.
In the context of this exercise, our goal is to use the definite integral to sum the volumes of cylindrical shells to get the total volume of the solid formed by revolving the region around the y-axis.
The general process involves:

• Identifying the appropriate limits of integration to cover the extent of the region. In our case, these are from \(y = 0\) to \(y = 9\).
• Expressing the radius and height of each shell in terms of a single variable, which is done using the equations bounding the region.

The definite integral \[ \int_{a}^{b} y^{3} \, dy \] is evaluated by finding the antiderivative and then computing the difference between the antiderivative's values at the upper and lower limits, \(b\) and \(a\). This computation results in the total volume of the solid.

Revolving Around y-axis

Revolving a region around an axis creates a three-dimensional solid. When the revolution is around the y-axis, it means you're spinning the two-dimensional region around the vertical axis to form a solid. This rotation turns vertical slices of the region into cylindrical shells, which are the basis for the previously discussed method.
To revolve around the y-axis, we typically consider regions bounded by functions expressed as x in terms of y. This is slightly different from revolving around the x-axis, where functions typically express y in terms of x.
For our problem:

• We identified the revolves region, initially defined in terms of y from the equation \(x = y^{3/2}\) .
• The shell's height is defined by how far the curve extends horizontally for each y-value, which is crucial when set up with the integral finding volume.

Revolving around the y-axis often requires a thoughtful setup of integrals but is an immensely powerful technique, especially for handling asymmetric regions. By visualizing the revolution process, understanding the solid's final shape becomes clearer, which can aid in setting up the correct integral.