Problem 22 - Calculus Single Variable

Question

In each of Exercises 19-24, use the method of washers to calculate the volume \(V\) obtained by rotating the given planar region \(\mathcal{R}\) about the \(y\) -axis. \(\mathcal{R}\) is the region between the curves \(y=x-1\) and \(y=\log _{2}(x), 1 \leq x \leq 2\)

Step-by-Step Solution

Answer

The volume \( V \) is \( \pi \left[ \frac{7}{3} - \frac{4}{\ln(4)} \right] \).

Step 1: Identify the Curves and Intersection Points

The curves provided are \( y = x - 1 \) and \( y = \log_2(x) \). To rotate around the \( y \)-axis, we'll first ensure the region is well-defined by finding the points where these curves intersect for the interval \( 1 \leq x \leq 2 \). Solve \( x - 1 = \log_2(x) \) to find intersection points. Numerical methods or plotting shows they don't intersect within the interval; hence we use the given bounds.

Step 2: Set Up the Washer Formula

The volume \( V \) of the solid formed by rotating about the \( y \)-axis is calculated using washers. The washer formula is \( V = \pi \int_{a}^{b} [R(y)^2 - r(y)^2] \, dy \), where \( R(y) \) and \( r(y) \) are the outer and inner radii of the washers, respectively.

Step 3: Express the Functions in terms of \(y\)

For rotation around the \( y \)-axis, rewrite joints in terms of \( y \). Solve \( y = x - 1 \) for \( x \) gives \( x = y + 1 \). Solve \( y = \log_2(x) \) for \( x \) gives \( x = 2^y \). Determine which is the outer radius \( R(y) = y + 1 \) and inner radius \( r(y) = 2^y \) since \( y + 1 \) is further right for \( y \)-axis.

Step 4: Determine the Integration Limits in terms of \( y \)

Convert the given \( x \)-interval \( 1 \leq x \leq 2 \) into \( y \)-values. For \( y = \log_2(x) \), the range for \( y \) is from \( \log_2(1) = 0 \) to \( \log_2(2) = 1 \) corresponding to \( x = 1 \) and \( x = 2 \).

Step 5: Compute the Integral

Substitute \( R(y) = y + 1 \) and \( r(y) = 2^y \) into the washer formula: \[ V = \pi \int_{0}^{1} [(y + 1)^2 - (2^y)^2] \, dy \]This integral is calculated by expanding both terms and integrating each:\[ (y + 1)^2 = y^2 + 2y + 1 \] \[ (2^y)^2 = 4^y \]Hence,\[ V = \pi \int_{0}^{1} (y^2 + 2y + 1 - 4^y) \, dy \]Evaluate this to find the volume.

Step 6: Evaluate the Resulting Integral

Integrate term by term:\[ \int (y^2 + 2y + 1) \, dy = \frac{y^3}{3} + y^2 + y \] \[ \int 4^y \, dy = \frac{4^y}{\ln(4)} \] (using exponential integration).Thus,\[ V = \pi \left[ \left( \frac{y^3}{3} + y^2 + y \right) - \frac{4^y}{\ln(4)} \right]_{0}^{1} \]Evaluate at the limits 0 and 1 for the result:\[ = \pi \left[ \left( \frac{1^3}{3} + 1^2 + 1 \right) - \frac{4^1}{\ln(4)} - \left( \frac{0^3}{3} + 0^2 + 0 \right) + \frac{4^0}{\ln(4)} \right] \]

Step 7: Calculate & Simplify Expression

Insert the evaluated expressions:\[ V = \pi \left[ \left( \frac{1}{3} + 1 + 1 \right) - \frac{4}{\ln(4)} \right] \]\[ = \pi \left[ \frac{7}{3} - \frac{4}{\ln(4)} \right] \]This gives the precise volume after simplification.

Key Concepts

Washer Method

The Washer Method is an essential technique in calculus used to find the volume of a solid of revolution. It's primarily applicable for regions bounded by two curves, which, when rotated around an axis, form a hollow object. Imagine a stack of washers, each with a hole in the middle—these washers help represent the volume of the solid that's created by the rotation of the region.

To apply the Washer Method, here's how you can start:

• Identify which of the given curves acts as an outer boundary and which one serves as the inner boundary. These curves determine the "outer" and "inner" radii for the washers.
• The difference in these radii gives the thickness of each slice, which, when integrated, calculates the total volume.

The formula for the volume using the Washer Method is given by:\[ V = \pi \int_{a}^{b} [R(y)^2 - r(y)^2] \, dy \]This expression represents the volume where the outer radius is \( R(y) \) and the inner radius is \( r(y) \), with the limits of integration from \( a \) to \( b \). It's crucial to express both radii in terms of the variable of integration, whether \( x \) or \( y \).

Rotating Around y-Axis

Rotating a region around the \( y \)-axis is a common problem in calculus that requires a solid understanding of setting up the appropriate radii for integration. When you rotate the region around the \( y \)-axis, you're essentially creating a 3D object from a 2D plane. Visualize this as revolving a flat shape around a central vertical line which adds depth to the original shape.

In this context, understanding the positioning of your curves is critical:

• You need to express the initial functions in terms of \( y \) since rotation is around the vertical \( y \)-axis.
• This change of perspective addresses the alignment of the curves with respect to the axis of rotation.

By examining whether each curve serves as an outer or inner boundary, you then adjust for the radii accordingly. This ensures that when you apply the Washer Method, the calculation respects the geometry formed by the rotation.

Integrating in Terms of y

Switching from integrating in terms of \( x \) to \( y \) may seem complex but is essential when dealing with problems involving vertical rotations. By expressing functions and setting up the integration in terms of \( y \), the setup aligns with the geometry of the rotation.

Here's how you handle this process:

• First, solve each given function to express \( x \) in terms of \( y \). This can involve algebraic manipulation or, in some cases, understanding the properties of the functions, such as inverses or logarithms.
• Once the functions are rewritten, evaluate which function extends further from the axis of rotation to define the outer boundary.
• The integration limits must also be adjusted to reflect the corresponding \( y \)-values. This involves calculating the range of \( y \) from the initial bounds of \( x \).

By mastering this transition, you ensure smoother and more accurate calculations.

Logarithmic Functions

Logarithmic functions often appear in calculus problems that involve volumes and areas, adding a layer of complexity due to their unique properties. The logarithm \( y = \log_2(x) \) represents one of the curves in our problem, and understanding it is crucial to accurately determine the volume.

Handling logarithmic functions during such calculations involves:

• Recognizing their typical shape—a curve that increases but at a decreasing rate, flattens as \( x \) increases.
• Algebraically solving the logarithmic equation to express one variable in terms of another. Here, it involves converting \( y = \log_2(x) \) to \( x = 2^y \). This conversion is vital since calculating volume requires expressing the radii in terms of the rotated axis (in this case, \( y \)).

Additionally, when integrating logarithmic functions, exponential properties play a central role. In the formula \( \int 4^y \, dy = \frac{4^y}{\ln(4)} \), it illustrates how understanding the derivative and integral relations of logs and exponents is pivotal. Such properties allow the calculation of definite integrals, crucial for determining areas and volumes.