Question

For the following regions \(R\), determine which is greater - the volume of the solid generated when \(R\) is revolved about the \(x\) -axis or about the y-axis. \(R\) is bounded by \(y=1-x^{3}\), the \(x\) -axis, and the \(y\) -axis.

Short answer

Answer: The volume generated when revolving region R about the x-axis is greater.

Step-by-step solution

Find the intersection points

To find the intersection points of \(y=1-x^3\) with the \(x\) and \(y\) axes, we need to solve the equations for the two axes: 1. For the \(x\)-axis, set \(y = 0\) and solve for \(x\): \(0 = 1 - x^3\). Let's find the value for \(x\): \(x^3 = 1 \Rightarrow x = 1\) 2. For the \(y\)-axis, set \(x = 0\) and solve for \(y\): \(y = 1 - (0)^3 = 1\) So, the region \(R\) is enclosed by the curve \(y=1-x^3\), the \(x\)-axis, and the \(y\)-axis within the interval \(0 \le x \le 1\).

Determine the volume when \(R\) is revolved about the \(x\)-axis

Let's use the disk method to find the volume when region \(R\) is revolved about the \(x\)-axis. The formula for the volume generated by the disk method is given by \(V = \pi \int{y^2\:dx}\). First, we must solve for \(y^2\), \(y^2 = (1 - x^3)^2 = 1 - 2x^3 + x^6\). Now, let's integrate \(y^2\) from to \(1\): \(V_x = \pi \int_{0}^{1} (1 - 2x^3 + x^6)\: dx = \pi\left[ x - \frac{1}{2}x^4 + \frac{1}{7}x^7\right]_0^1\) Plugging in the bounds of integration: \(V_x = \pi\left(1 - \frac{1}{2} + \frac{1}{7}\right) - \pi\left(0\right) = \pi \left(\frac{9}{14}\right)\)

Determine the volume when \(R\) is revolved about the \(y\)-axis

Using the disk method, we need to solve for \(x\) in terms of \(y\): \(x = \sqrt[3]{(1 - y)}\) Now, square the expression of \(x\): \(x^2 = (1 - y)^{\frac{2}{3}}\) To use the disk method, integrate the above expression with respect to \(y\) from \(0\) to \(1\): \(V_y = \pi \int_{0}^{1} (1 - y)^{\frac{2}{3}}\: dy\) To evaluate this integral, we can perform a substitution, Let \(u = 1 - y\), then \(-du = dy \Rightarrow dy = -du\) and the integration limits change to \(u=1\) and \(u=0\). So, we get, \(V_y = -\pi \int_{1}^{0} u^{\frac{2}{3}}\: du\) Changing the limits of integration and removing the negative sign: \(V_y = \pi \int_{0}^{1} u^{\frac{2}{3}}\: du = \pi\left[\frac{3}{5}u^{\frac{5}{3}}\right]_0^1\) Plugging in the bounds of integration: \(V_y = \pi\left(\frac{3}{5}\right) - \pi\left(0\right) = \pi\left(\frac{3}{5}\right)\)

Compare the volumes

Now that we have calculated the volumes generated when rotating region \(R\) about the \(x\)-axis and the \(y\)-axis, we need to compare them to determine which is greater: \(V_x = \pi\left(\frac{9}{14}\right)\) \(V_y = \pi\left(\frac{3}{5}\right)\) Since \(\frac{9}{14}\) is greater than \(\frac{3}{5}\), we conclude that the volume generated when revolving region \(R\) about the \(x\)-axis is greater.

Key concepts

Disk Method

The disk method is a technique in calculus used to find the volumes of solids of revolution. Imagine taking a flat region and revolving it around an axis. The solid formed is like a stack of disks (or thin cylinders). Each disk's volume can be approximated with the formula for the volume of a cylinder, using the formula \ \( V = \pi r^2 h \ \), where \ \( r \) is the radius and \ \( h \) the thickness.

**Steps to Use the Disk Method:**

• Identify the axis of rotation.
• Determine the radius of each disk. This is typically the distance from the curve to the axis of rotation.
• Integrate the area of the disks over the interval.

For example, when revolving a region around the x-axis, your radius is typically the y-value of the function, and you'll integrate with respect to x. If you revolve around the y-axis, you solve for x in terms of y and integrate with respect to y.

Integration

Integration is a fundamental concept of calculus that helps to accumulate quantities, such as area under curves or volume of solids. Think of integration as the opposite of differentiation. It combines small parts to find a total.

**How Integration Works in Disk Method:**

• You first find the expression for the radius of the disks.
• Square this function since the volume of each disk involves the area of a circle.
• The integral then sums the volumes of these disks over a specific interval.

For example, to find the volume when revolving a function around the x-axis, integrate \ \( y^2 \) within your interval of interest. Each elemental disk contributes a small volume which, when summed through integration, gives the total volume of the solid.

Volumes of Solids of Revolution

Volumes of solids of revolution are calculated by revolving a 2D area around an axis to create a 3D object. This is a great way to determine volumes of complex shapes that aren't easily calculated with basic geometry.

When considering solids of revolution, it's important to decide which axis will be the center of rotation, as this affects the shape and therefore the volume of the solid.

**Steps to Determine Volume:**

• Define the boundaries of the region being revolved. In our example, it's the area between the curve and the axes.
• Choose the correct integration method, like the disk method, based on the axis of rotation and shape boundaries.
• Apply integration to compute the total volume.

This method is vital, especially in engineering and physical sciences, where creating physical objects and determining material usage and structural integrity is necessary.