Problem 10 - Calculus Single Variable

Question

In each of Exercises 7-12, use the method of disks to calculate the volume \(V\) of the solid that is obtained by rotating the given planar region \(\mathcal{R}\) about the \(y\) -axis. \(\mathcal{R}\) is the region that is bounded on the left by the \(y\) -axis, on the right by the curve \(y=x^{2}, 1 \leq x \leq 2,\) and that is between the horizontal lines \(y=1\) and \(y=4\).

Step-by-Step Solution

Answer

The volume of the solid is \( \frac{15\pi}{2} \).

Step 1: Identify the Region

The region \( \mathcal{R} \) is bounded by the curve \( y = x^2 \), the vertical line \( x = 2 \), the horizontal line \( y = 4 \), and the \( y \)-axis, which corresponds to \( x = 0 \). The range for \( x \) is from 1 to 2.

Step 2: Set up the Integral

When rotating around the \( y \)-axis using the disk method, we use the formula for the volume: \[ V = \pi \int_{a}^{b} [R(y)]^2 \, dy \]. The function \( x = \sqrt{y} \) gives the radius of the disks. The limits for \( y \) are 1 to 4.

Step 3: Express the Radius Function

The radius \( R(y) \) is the distance from the \( y \)-axis to the function \( x = \sqrt{y} \). Thus, \( R(y) = \sqrt{y} \).

Step 4: Calculate the Volume Integral

Substitute \( R(y) = \sqrt{y} \) into the integral. Thus, \[ V = \pi \int_{1}^{4} (\sqrt{y})^2 \, dy = \pi \int_{1}^{4} y \, dy \].

Step 5: Evaluate the Integral

Compute the integral: \[ V = \pi \left[ \frac{y^2}{2} \right]_{1}^{4} = \pi \left( \frac{16}{2} - \frac{1}{2} \right) = \pi \left( 8 - \frac{1}{2} \right) = \pi \times \frac{15}{2} \].

Step 6: Calculate the Final Volume

Simplify the expression: \[ V = \frac{15\pi}{2} \]. Therefore, the volume of the solid is \( \frac{15\pi}{2} \).

Key Concepts

Volume of Solid of Revolution

The volume of a solid of revolution is a concept in calculus where a region in the plane is revolved around a line (often an axis), creating a three-dimensional object. To find the volume of such objects, we use techniques from integral calculus. One effective method is the disk method.
This involves slicing the solid into numerous tiny disk shapes along the axis of rotation. By calculating the volume of each small disk and summing them up, we can estimate the entire volume of the solid.

• The disks are thin cylinders with radius determined by the distance from the axis of revolution to the edge of the region.
• The height of each disk is an infinitesimally small thickness, typically expressed as a differential, like \(dy\) or \(dx\).

In our exercise, the region is rotated around the y-axis, generating disks perpendicular to this axis.

Integral Calculus

Integral Calculus plays a crucial role in finding the volume of solids of revolution. By using integration, we can sum up an infinite number of infinitesimally small elements to find total quantities such as area and volume.
In this context, the integral calculates the summation of the areas of the disks throughout the interval of the region. The formula for volume of revolution using the disk method is:
\[ V = \pi \int_{a}^{b} [R(y)]^2 \, dy \]
Here, \(R(y)\) represents the radius function, and \(a\) and \(b\) are the limits of integration corresponding to the boundaries of the region along the axis of rotation.

• The function \(R(y)\) needs to be expressed in terms of the variable of integration (here \(y\)), which ensures we are integrating along the correct axis.
• The complete integral provides an exact measure of the volume of the solid.

Rotating About the y-axis

When rotating a region about the y-axis, it's essential to visualize how the original two-dimensional shape evolves.
Instead of horizontal cross-sections, as when rotating around the x-axis, the slices here are vertical, making circles or disks parallel to the yz-plane.

• The radius of each disk at a given height \(y\) is determined by the horizontal distance from the y-axis to the curve, which in this case is \(x = \sqrt{y}\).
• Therefore, the radii of the disks change as you move along the y-axis, from the smallest at \(y = 1\) to the largest at \(y = 4\).

Visualizing this helps in setting up the correct function and limits for the integral calculation.

Integral Evaluation

The final step in finding the volume is evaluating the integral set up in the previous steps.
Once you've established the integral with the volume formula, you carry out the integration process to solve it.
In our exercise:

• The integral \[ V = \pi \int_{1}^{4} y \, dy \] is computed by finding the antiderivative of \(y\).
• The result is \[ \pi \left[ \frac{y^2}{2} \right]_{1}^{4} \]
• Evaluating from 1 to 4 gives \[ \pi \left( \frac{16}{2} - \frac{1}{2} \right) = \pi \times \frac{15}{2} \]

This systematic evaluation provides the precise volume of the solid, demonstrating the power of integral calculus in practical applications.