Problem 26 - Calculus an Applied Approach

Question

Use a double integral to find the volume of the solid bounded by the graphs of the equations. $$ z=x+y, x^{2}+y^{2}=4 \text { (first octant) } $$

Step-by-Step Solution

Answer

The volume of the solid bounded by the graph of the equations is \(\frac{16}{3}\).

Step 1: Convert to Polar Coordinates

For the circle equation \(x^2 + y^2 = 4\), we convert these Cartesian coordinates to polar coordinates. We use \(x = r \cos(\theta)\) and \(y = r \sin(\theta)\). The circle equation can now be expressed as \(r^2 = 4\), which simplifies to \(r=2\) since we're only considering the first octant where \(r \geq 0\). The volume will be integrated over the quarter-circle sector of a polar coordinate system, that is, \(\theta\) will range from 0 to \(\frac{\pi}{2}\).

Step 2: Set up the Double Integral

We rewrite \(z = x + y\) using polar coordinates to get \( z = r \cos(\theta) + r \sin(\theta) = r(\cos(\theta) + \sin(\theta))\). The volume of the solid projected onto the quarter circle disk in the xy-plane will be given by the double integral \(V = \int \int z \, dA\). Using \(\,dA = r \, dr \, d\theta\) in polar coordinates, the integral becomes \(V = \int_{0}^{\frac{\pi}{2}} \int_{0}^{2} r^2 (\cos(\theta) + \sin(\theta)) \, dr \, d\theta\).

Step 3: Evaluate the Double Integral

We first integrate with respect to \(r\): ∫ from 0 to 2 of \(r^2 (\cos(\theta) + \sin(\theta)) \, dr = \frac{1}{3} r^3 (\cos(\theta) + \sin(\theta)) ∣ from 0 to 2 = \frac{8}{3} (\cos(\theta) + \sin(\theta))\). Then integrate this result from 0 to \(\frac{\pi}{2}\) with respect to \(\theta\): ∫ from 0 to \(\frac{\pi}{2}\) of \(\frac{8}{3} (\cos(\theta) + \sin(\theta)) \, d\theta = \frac{8}{3}(\sin(\theta) - \cos(\theta)) ∣ from 0 to \(\frac{\pi}{2}\) = \frac{16}{3}\).

Step 4: Conclusion

The volume of the solid bounded by the graph of the given equations is \(\frac{16}{3}\).

Key Concepts

Polar Coordinates

Polar coordinates provide a different way to represent a point in a plane, offering an alternative to the traditional Cartesian (x, y) system. This system is particularly useful when dealing with problems that are symmetrical around a point, like circles or sectors. There are two main components in polar coordinates: the radial distance \(r\) and the angle \(\theta\). Here’s a simple breakdown:

• Radial distance \(r\): This is the distance from the origin to the point. In our exercise, it's derived from the circle equation \(x^2 + y^2 = 4\), simplifying to \(r=2\) for the given circle.
• Angle \(\theta\): This angle is measured from the positive x-axis. For a quarter-circle sector in the first octant, \(\theta\) ranges from 0 to \(\frac{\pi}{2}\).

By expressing coordinates in this way, we accommodate rotational symmetry more easily. This greatly simplifies the integration processes when dealing with circular or sector shapes, such as in our exercise.

Volume of Solid

Finding the volume of a solid using integrals is a common task in calculus, and using double integrals with polar coordinates can simplify the process significantly. Let's delve into each part.To determine the volume of the solid, we need to evaluate the double integral \(V = \int \int z \, dA\). In polar coordinates, \(dA\) becomes \(r \, dr \, d\theta\), accounting for the circular nature of the plane. Given the equation \(z = x + y\) translates into \(z = r(\cos(\theta) + \sin(\theta))\) in polar form:

• First, integrate with respect to \(r\), which accounts for radial distance, from 0 to 2.
• Then, integrate with respect to \(\theta\), which accounts for the angle, from 0 to \(\frac{\pi}{2}\).

By breaking down into these steps, we find the total volume over the specified sector, ensuring each layer from the origin to the boundary is evaluated. This focus on the radial sweep of \(r\) and the angle \(\theta\) elegantly tackles the symmetrical nature of our shape.

Quarter Circle Sector

The quarter circle sector is one-fourth of a circle, a shape we deal with frequently in polar coordinates. It's an important concept when considering regions confined to a specific part of the polar plane, like in this problem.In the exercise, we are interested in only one sector of a full circle. This involves:

• Sweeping \(r\), the radial line, from 0 to 2—half the full circle's radius in our setup.
• Adjusting \(\theta\), the angle, ranging from 0 to \(\frac{\pi}{2}\)—thus making it only a quarter of the full circle, hence the term "quarter circle sector."

Understanding the quarter circle sector gives clarity on why we choose specific \(r\) and \(\theta\) in the integration limits. It allows us to focus only on the part of the space that contributes to the total volume we compute. This careful consideration of angular and radial limits is essential in evaluating the integral correctly.