Problem 102 - Calculus Volume 1

Question

For the following exercises, draw the region bounded by the curves. Then, use the washer method to find the volume when the region is revolved around the \(y\) -axis. \(x=\sqrt{9-y^{2}}, x=e^{-y}, y=0,\) and \(y=3\)

Step-by-Step Solution

Answer

The volume is \( V = \pi (18 - \frac{1}{515}) \).

Step 1: Understand the Problem

We have to find the volume generated by rotating a bounded region around the y-axis. The region is enclosed by the curves \( x = \sqrt{9-y^2} \), \( x = e^{-y} \), \( y = 0 \), and \( y = 3 \).

Step 2: Sketch the Graph

Let's draw each curve in the xy-plane. \( x = \sqrt{9-y^2} \) is the right half of the circle with radius 3 centered at the origin. \( x = e^{-y} \) is an exponential decay function. The lines \( y = 0 \) (the x-axis) and \( y = 3 \) are horizontal lines.

Step 3: Identify the Bounded Region

The region of interest is where the curves \( x = \sqrt{9-y^2} \) and \( x = e^{-y} \) intersect between \( y = 0 \) and \( y = 3 \). This is the area beneath \( x = e^{-y} \) and up to the circle.

Step 4: Set Up the Washer Method

The volume of revolution around the y-axis is found using washers. The outer radius is given by \( R(y) = \sqrt{9-y^2} \) and the inner radius by \( r(y) = e^{-y} \).

Step 5: Volume Integral Setup

The formula for the volume using the washer method is \[ V = \pi \int_{a}^{b} \left( R(y)^2 - r(y)^2 \right) \, dy \]Substituting the functions, the integral becomes \[ V = \pi \int_{0}^{3} \left( (\sqrt{9-y^2})^2 - (e^{-y})^2 \right) \, dy \] which simplifies to \[ V = \pi \int_{0}^{3} \left( 9-y^2 - e^{-2y} \right) \, dy \]

Step 6: Evaluate the Integral

Calculate the integral over the interval from \( y = 0 \) to \( y = 3 \):\[ V = \pi \left[ (9y - \frac{y^3}{3} - \frac{e^{-2y}}{2}) \right]_{0}^{3} \]Substitute the limits and evaluate:\[ V = \pi \left[ (27 - 9 - \frac{1}{2}e^{-6}) \right] - \pi \left[ (0 - 0 - \frac{1}{2}) \right] \]

Step 7: Simplifying the Result

The volume calculation leads to:\[ V = \pi (27 - 9 - \frac{1}{515}) \] Which equals \[ V = 18\pi - \frac{1}{515}\pi \].

Step 8: Final Volume Result

The final volume of the solid formed is \[ V = \pi (18 - \frac{1}{515})\].

Key Concepts

Calculus

Calculus is a powerful branch of mathematics that helps us understand change and motion. It provides tools to calculate things like rates of change, lengths, areas, and volumes.
For solving real-world problems, calculus is indispensable. In the context of the washer method, calculus allows us to compute complex volumes with ease.
Here are some core ideas in calculus:

• **Derivatives:** These help us understand the rate at which something changes. Think of it as the gradient or slope at any point on a curve.
• **Integrals:** Contrary to derivatives, integrals help us sum up small parts to find totals, like capturing the total area underneath a curve.
• **Limits:** The foundation of calculus, limits help define both derivatives and integrals by considering what happens as we approach specific points.

Volume of Revolution

When tackling the "Volume of Revolution", it's about revolving a shape around an axis to create a 3-dimensional object. Imagine rotating a 2D region, like a circle, around an axis; it forms a solid object, such as a sphere.
**Washer Method**The Washer Method is a technique in calculus used specifically for finding volumes of 3D objects that have holes or are hollow. It gets its name from the idea of stacking washers—thin discs with holes in their centers.
Here's how the method functions:

• The `washer` is a thin slice of the object perpendicular to the axis of revolution.
• Each washer has an `outer radius` and an `inner radius`.
• The volume of each washer is calculated using the formula for area of a circle \(\pi(R^2-r^2)\) times its thickness (usually an axis increment).

This method is particularly useful when the outer radius and inner radius of the washers change, allowing for the accurate calculation of more complex shapes.

Integral Calculus

Integral calculus deals primarily with calculating totals and accumulations, such as areas under curves and volumes. It's about bringing together all the small pieces.
The Washer Method makes extensive use of integral calculus because calculating the overall volume involves summing the volumes of infinite washers.
Here’s the general approach applied here:

• Identify the curves that bound the region of interest.
• Set up an integral to represent the total volume, from the start to the end of the bounded region (here, from \(0\) to \(3\)).
• Integrate the expression \((\pi\int_{a}^{b}(R(y)^2-r(y)^2)\, dy)\) to find volume.

Through integrating, we essentially sum up all infinitesimally small washer volumes to form a solid figure.
Integral calculus empowers us to solve such problems with precision.