Question

Use the shell method to set up and evaluate the integral that gives the volume of the solid generated by revolving the plane region about the \(y\) -axis. $$ y=\frac{1}{2} x^{2}, \quad y=0, \quad x=6 $$

Short answer

The volume of the solid of revolution is \(324\pi\) cubic units or approximately 1017.876 cubic units.

Step-by-step solution

Set up the integral using the shell method

When using the shell method, the shell's volume is given by the formula \(2π * x * height * thickness\), which in this case is \(2π * x * y * dx\). As \(y = \frac{1}{2} x^{2}\), the volume of the shell is \(2π * x * \frac{1}{2} x^{2} * dx\), or \(π * x^{3} * dx\). The integral is found by evaluating this expression from \(x = 0\) to \(x = 6\). So the volume \(V\) is given by: \[V= \int_{0}^{6} \pi x^{3} dx\]

Evaluate the integral

The integral of \(x^{3}\) is \(\frac{1}{4} x^{4}\), so\[V=\pi \left[ \frac{1}{4} x^{4} \right]_{0}^{6} = \frac{\pi}{4} * 6^{4} - \frac{\pi}{4} * 0^{4}\]

Solve for the volume

Solving the equation above gives:\[V=\frac{6^{4}\pi}{4} = \frac{1296\pi}{4} = 324\pi \approx 1017.876 cubic units\]

Key concepts

Volume by Shell Method

Understanding the shell method is crucial for calculating the volume of solids of revolution, especially when the solid is rotated around an axis other than the axis of symmetry.

Imagine slicing a loaf of bread. Each slice represents a 'shell,' and when we stack these shells together, we form the solid. The shell method involves calculating the volume of each thin cylindrical shell and then integrating these volumes over the entire solid.

For a horizontal slice at a given height 'x', the radius of the shell is just 'x', its height is given by the function value, such as our example \( y = \frac{1}{2}x^2 \) and the thickness is an infinitesimally small \( dx \). The formula for the volume of a single shell becomes \( 2\pi * x * height * thickness \), or \( 2\pi * x * y * dx \) for our exercise.

By integrating this expression between the bounds of the solid (from \( x=0 \) to \( x=6 \) in our exercise), we sum up all the shells' volumes to find the total volume of the solid.

Solid of Revolution

A solid of revolution is a three-dimensional object created by rotating a two-dimensional plane area around a line (the axis of revolution). This process is akin to turning pottery on a wheel, where a shape, such as a curve or rectangle, is spun around to form a symmetrical solid.

In our exercise, the region between the curve \( y=\frac{1}{2} x^2 \) and the lines \( y=0 \) and \( x=6 \) is rotated about the y-axis to form a paraboloid.

Visualizing the Revolution

Imagine the graph of the curve extending upwards from the x-y plane and sweeping around the y-axis. Each point on the curve traces a circle whose radius equals the x-coordinate of that point. The complete set of these circles forms the outer boundary of the solid, generating a shape somewhat like a flared vase or bell.

Definite Integral

The definite integral is a cornerstone of calculus, used to calculate areas, volumes, and total quantities where the rate of change is known. When it comes to finding the volume of a solid of revolution using the shell method, the definite integral accumulates the volumes of infinitesimally thin shells across the solid.

In our problem, the volume \( V \) is given by the definite integral of \( \pi x^3 \) from \( x=0 \) to \( x=6 \) which symbolically captures the sum of all cylindrical shells' volumes to find the total volume of the 3D shape.

Evaluating the Integral

The integral \( \int_{0}^{6} \pi x^{3} dx \) is evaluated using the power rule for integration, giving us \( \frac{1}{4} x^{4} \) when applied to \( x^3 \). With our limits, we plug in the upper bound \( x=6 \) and subtract the volume evaluated at the lower bound \( x=0 \) to find the definite volume of our solid.