Problem 62 - University Calculus: Early Transcendentals

Question

Find the volume of the region bounded above by the sphere \(x^{2}+y^{2}+z^{2}=2\) and below by the paraboloid \(z=x^{2}+y^{2}\).

Step-by-Step Solution

Answer

The volume is \( \frac{\pi}{3} \).

Step 1: Identify Intersections

To determine the volume between the sphere and the paraboloid, first identify where these two surfaces intersect. Set the equations equal to each other: \[ x^2 + y^2 + (x^2 + y^2)^2 = 2. \] Simplifying, we have:\[ 2x^2 + 2y^2 = 2 \] which leads to \[ x^2 + y^2 = 1. \] This equation describes a circle of radius 1 in the xy-plane.

Step 2: Determine Limits of Integration

The intersection in the xy-plane, given by the circle \( x^2 + y^2 = 1 \), tells us that the limits of integration in polar coordinates will be \( 0 \leq r \leq 1 \) and \( 0 \leq \theta \leq 2\pi \) for the radius and angle.

Step 3: Set Up Volume Integral

Now, calculate the volume with the region above the paraboloid and below the sphere. In polar coordinates, we express the equations as:- Sphere: \( z = \sqrt{2 - x^2 - y^2} = \sqrt{2 - r^2} \)- Paraboloid: \( z = r^2 \)The volume integral is:\[ V = \int_{0}^{2\pi} \int_{0}^{1} \left( \sqrt{2 - r^2} - r^2 \right) r \, dr \, d\theta \]

Step 4: Evaluate Integral

Evaluate the integral with respect to \( r \):\[ \int_{0}^{1} \left( \sqrt{2 - r^2} - r^2 \right) r \, dr \]Split into two integrals:\[ \int_{0}^{1} r \sqrt{2 - r^2} \, dr - \int_{0}^{1} r^3 \, dr. \]- For \( \int_{0}^{1} r \sqrt{2 - r^2} \, dr \), use substitution \( u = 2 - r^2 \), \( du = -2r \, dr \).- For \( \int_{0}^{1} r^3 \, dr \), simply integrate directly: \( \frac{r^4}{4} \bigg|_{0}^{1}. \)Compute these integrals separately and then subtract the results: \[ \frac{-1}{3} \left( (2 - (0)^2)^{3/2} - 2^{3/2} \right) - \frac{1}{4}. \]

Step 5: Compute Total Volume

Integrate over \( \theta \) to find the total volume:\[ V = \int_{0}^{2\pi} \left( \frac{-1}{3} [1] + \frac{1}{4} \right) \, d\theta = \left( \frac{1}{6} \right) 2\pi. \]Thus, compute:\[ V = \frac{\pi}{3}. \]

Key Concepts

Spherical Coordinates

The concept of spherical coordinates might appear daunting at first, but their utility in solving problems involving spheres makes them invaluable. In Cartesian coordinates, a point in 3D space is given by three variables: \( (x, y, z) \). Spherical coordinates, however, use radius \( r \), inclination \( \theta \), and azimuth \( \phi \). This conversion is particularly useful when dealing with problems involving spherical symmetry:

• \( r \): The distance from the origin to the point.
• \( \theta \): The angle between the positive z-axis and the line connecting the origin to the point.
• \( \phi \): The angle from the positive x-axis to the projection of the point on the xy-plane.

By using spherical coordinates, we can greatly simplify the mathematical manipulations needed when integrating to find volumes or areas of spheres and spherical regions.

Polar Coordinates

Polar coordinates transform problems dealing with circles into much simpler forms compared to Cartesian coordinates. In this system, each point is described not by an \( x \) and \( y \) position, but by:

• \( r \): The distance from the origin to the point.
• \( \theta \): The angle from the positive x-axis to the point.

Using polar coordinates in integration allows us to leverage the inherent symmetry of circular shapes. For example, in our original problem, the intersection of the sphere and paraboloid produced a circle \( x^2 + y^2 = 1 \). This suggests that converting to polar coordinates will streamline the solution, as it aligns with the geometry of the problem. The limits of integration naturally become \( 0 \leq r \leq 1 \) and \( 0 \leq \theta \leq 2\pi \), outlining the circular region we need to evaluate.

Integration Techniques

Integration is a powerful tool in calculus that lets us determine quantities like area, volume, and other physical properties. There are key techniques to master, each applicable in different scenarios.For the exercise at hand, two important techniques were used:

• Substitution: A method used to simplify integrals by changing variables, making a complex integral easier to evaluate. In this case, substitution was applied to transform the integrand \( r \sqrt{2 - r^2} \) into a form more amenable to calculation.
• Splitting Integrals: When you have a sum or difference inside an integral, you can split the integrals apart. This separation makes evaluating each part more straightforward, especially when the pieces simplify to basic forms.

Mastering these techniques equips you to tackle a wide range of integration problems involving volumes or areas in different coordinate systems.

Intersection of Surfaces

In multivariable calculus, intersections of surfaces can represent interesting geometric shapes, crucial in determining regions to integrate over. To find where two surfaces intersect, we set their equations equal.In the original problem, the intersection of a sphere \( x^2 + y^2 + z^2 = 2 \) and a paraboloid \( z = x^2 + y^2 \) was crucial for setting up the bounds for integration. By equaling the expressions for \( z \), we solved for \( x^2 + y^2 = 1 \), revealing the circular intersection in the xy-plane.Understanding these intersections is pivotal:

• They define the region of space we are evaluating.
• They set the limits for our integrals, crucial for accuracy in the final calculation.

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