Problem 53 - Advanced Engineering Mathematics

Question

In Problems \(51-54\), find the volume of the solid that is bounded by the graphs of the given equations. $$ z=x^{2}+y^{2}, x^{2}+y^{2}=25, \quad z=0 $$

Step-by-Step Solution

Answer

The volume of the solid is \( \frac{625 \pi}{2} \).

Step 1: Understand the Problem

The problem asks to find the volume of the solid bounded by the surfaces. The given equations represent a paraboloid \( z = x^2 + y^2 \) and a cylinder \( x^2 + y^2 = 25 \) slicing the paraboloid with a base at \( z = 0 \).

Step 2: Set up the Integral

To find the volume beneath the paraboloid, we integrate over the circular base defined by \( x^2 + y^2 \leq 25 \). The height at each point \((x, y)\) is \( z = x^2 + y^2 \). Thus, the volume \( V \) is given by the double integral: \ \[ V = \int_{-5}^{5} \int_{-\sqrt{25-x^2}}^{\sqrt{25-x^2}} (x^2 + y^2) \, dy \, dx \]

Step 3: Convert to Polar Coordinates

This problem is more suited for polar coordinates, where \( x = r \cos \theta \) and \( y = r \sin \theta \). So, \( x^2 + y^2 = r^2 \) and the bounds become \( 0 \leq r \leq 5 \) and \( 0 \leq \theta \leq 2\pi \). The integral becomes: \ \[ V = \int_{0}^{2\pi} \int_{0}^{5} r^2 \cdot r \, dr \, d\theta \] \ Simplifying gives: \ \[ V = \int_{0}^{2\pi} \int_{0}^{5} r^3 \, dr \, d\theta \]

Step 4: Integrate with Respect to \( r \)

Perform the integration with respect to \( r \): \ \[ V = \int_{0}^{2\pi} \left[ \frac{r^4}{4} \right]_{0}^{5} \, d\theta = \int_{0}^{2\pi} \frac{625}{4} \, d\theta \]

Step 5: Integrate with Respect to \( \theta \)

Now, integrate the constant with respect to \( \theta \): \ \[ V = \left[ \frac{625}{4} \cdot \theta \right]_{0}^{2\pi} = \frac{625}{4} \times 2\pi \] \ Evaluate this to get the final volume: \ \[ V = \frac{625 \pi}{2} \]

Step 6: Conclusion

So, the volume of the solid bounded by the given equations is \( \frac{625 \pi}{2} \).

Key Concepts

Double Integrals

Double integrals are a powerful mathematical tool used to calculate volumes, areas, and other properties over two-dimensional regions. They extend the concept of a single integral, which operates along a line, to operate over an area.
To evaluate a double integral, we integrate one variable while treating the other as a constant, and then integrate the resulting expression with respect to the second variable. This process is typically conducted over a defined region in the xy-plane.

• In the context of this problem, the double integral calculates the volume beneath the paraboloid surface over a circular region defined by the cylinder's base.
• The double integral was initially expressed in rectangular coordinates as: \ \[ V = \int_{-5}^{5} \int_{-\sqrt{25-x^2}}^{\sqrt{25-x^2}} (x^2 + y^2) \, dy \, dx \]
• This integral sums small elements of volume in the region, which involve stacking up horizontal slices of the paraboloid from the base at \( z = 0 \) to its top limit.

Polar Coordinates

Polar coordinates are especially useful for problems with circular symmetry, such as those involving circles or spheres. Instead of using x and y coordinates, we use the radius \( r \) and the angle \( \theta \) to describe a point's location.
This shift from Cartesian coordinates (rectangular) to polar coordinates simplifies the integration process for shapes like circles.

• The conversion uses the relationships: \( x = r \cos \theta \) and \( y = r \sin \theta \).
• The differential area element \( dx \, dy \) in polar becomes \( r \, dr \, d\theta \), capturing both the radial and angular components.
• For this exercise, converting to polar coordinates simplified the bounds to \( 0 \leq r \leq 5 \) and \( 0 \leq \theta \leq 2\pi \), capturing the full circular base of the cylinder.

Paraboloid Surface

A paraboloid is a three-dimensional surface described by a quadratic equation, in this case, \( z = x^2 + y^2 \). Visually, it looks like a bowl or a dish.
In many mathematical problems, paraboloids represent surfaces of revolution obtained by rotating a parabola around its axis.

• The equation \( z = x^2 + y^2 \) defines a paraboloid opening upwards with its vertex at the origin \( (0,0,0) \).
• The height \( z \) at any point \( (x, y) \) is determined by the sum of the squares of \( x \) and \( y \), defining its sloped surface.
• In this problem, the paraboloid is sliced horizontally by \( z = 0 \) at its base, limiting the volume we calculate to the portion within the cylinder.

Cylindrical Coordinates

Cylindrical coordinates are a natural extension of polar coordinates into three dimensions, providing an efficient way to describe points with circular and linear features.
This coordinate system incorporates a height dimension to the radial and angular components of the polar system, making it ideal for dealing with problems involving cylinders.

• The coordinates \( (r, \theta, z) \) are used: \( r \) is the radial distance, \( \theta \) the angular coordinate, and \( z \) the height.
• A cylinder in this coordinate system is described with \( r = \text{constant} \). Here, the cylinder is portrayed by \( x^2 + y^2 = 25 \), or \( r = 5 \).
• This system simplifies the integration process by aligning one coordinate with the problem's symmetrical features, as seen when polar coordinates were utilized to integrate the paraboloid under the cylindrical bound.