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. \(x=y^{2 / 3}, y=27, x=0\)

Short answer

The volume of the solid is \(\frac{2187\pi}{7}\).

Step-by-step solution

Understand the Boundary

The region is bounded by the curves \(x = y^{2/3}\), \(y = 27\), and \(x = 0\). This means the region is on the right side of the \(y\)-axis (\(x = 0\)), above \(y = 0\) and below \(y = 27\), and to the left of the curve \(x = y^{2/3}\).

Sketch the Region

Plot the curve \(x = y^{2/3}\), which is a sideways power function, starting from the origin. Then, draw the horizontal line \(y = 27\) across the \(x\)-axis and the vertical line \(x = 0\) (the \(y\)-axis). The region \(R\) is enclosed in the first quadrant between these curves.

Identify the Horizontal Slice

Imagine a horizontal slice of thickness \(dy\) at a certain \(y\). This slice extends from \(x = 0\) to \(x = y^{2/3}\) and has a width of \(y^{2/3}\). This slice will be revolved around the \(y\)-axis.

Set Up the Integral for Volume

The volume of the solid generated by revolving around the \(y\)-axis is found using the formula for the volume of revolution: \( V = \pi \int_{a}^{b} [f(y)]^2 \, dy \). Here, \(f(y) = y^{2/3}\), and \(a = 0\), \(b = 27\).

Substitute and Compute the Integral

Substitute \(f(y) = y^{2/3}\) into the integral to get \( V = \pi \int_{0}^{27} (y^{2/3})^2 \, dy \). Simplify to \( V = \pi \int_{0}^{27} y^{4/3} \, dy \).

Evaluate the Integral

To evaluate the integral, calculate the antiderivative of \(y^{4/3}\), which is \(\frac{3}{7} y^{7/3}\). So, the integral becomes \( \pi \left[ \frac{3}{7} y^{7/3} \right]_{0}^{27} \).

Final Calculation

Substitute the limits into the antiderivative: \( V = \pi \left[ \frac{3}{7} (27^{7/3}) - \frac{3}{7} (0^{7/3}) \right] = \pi \left( \frac{3}{7} \cdot 729 \right) = \pi \left( \frac{2187}{7} \right) \).

Simplify the Result

Calculate the exact numeric value: \( V = \frac{2187\pi}{7} \). Therefore, the volume of the solid is \(\frac{2187\pi}{7}\).

Key concepts

Integral Calculus and Volumes

Integral calculus is a powerful mathematical tool used to find areas, volumes, and other quantities that require summation over intervals. In the context of calculating volumes of revolution, it helps us determine the total volume of a solid obtained by rotating a region around a given axis. The volume

• is determined using integrals.
• requires a precise setup of the integral based on the axis of rotation.
• needs the correct bounds of integration to represent the region involved.

In this exercise, we compute the volume of a solid by rotating a region around the y-axis, using the formula for the volume of revolution: \[ V = \pi \int_{a}^{b} [f(y)]^2 \, dy \] This involves carefully setting up the integral with respect to the variable of the axis of rotation. Here we integrate with respect to \(y\), as it's rotated about the \(y\)-axis.

Understanding the Solid of Revolution

A solid of revolution is a three-dimensional object formed by rotating a two-dimensional area about an axis. This concept is essential for visualizing how a particular region generates a solid when revolved.

• The given region is bounded by specific curves or lines, such as \(x = y^{2/3}\), \(y = 27\), and \(x = 0\).
• The region is revolved around an axis—in this case, the y-axis—to create the solid.
• This process essentially "sweeps" the area around the axis, forming a symmetrical 3D shape.

In our specific problem, by rotating the region bounded by these functions, a volume is created that can be calculated precisely using integral calculus.

The Role of Antiderivatives

Antiderivatives, often referred to as indefinite integrals, are fundamental in solving integrals. They help us determine the accumulation of quantities, such as area or volume. To find the volume of a rotated solid, we will:

• Determine the antiderivative of the function inside the integral.
• Evaluate this antiderivative at the upper and lower bounds of the region.
• Subtract these evaluations to find the exact accumulated value (volume in this case).

For instance, when computing: \[ \int_{0}^{27} y^{4/3} \, dy \] the antiderivative is \( \frac{3}{7} y^{7/3} \). Calculating this at the bounds \(0\) and \(27\) gives us the total volume enclosed by the solid after rotation.

Identifying the Bounded Region

The bounded region is the specific area within the graph that we are interested in revolving around an axis. Identifying this region correctly is crucial:

• It defines the limits of integration \(a\) and \(b\).
• Provides a visual boundary for sketching the function curves, such as \(x = y^{2/3}\), \(y = 27\), and \(x = 0\).
• Needs accurate plotting to ensure precise computation of the integral.

In this problem, the region is bounded between these lines and curves, forming a distinct area to be rotated. Comprehending the boundary permits us to set up the integral accurately, ensuring that the calculated volume reflects the region's true rotation around the \(y\)-axis.