Problem 20 - Calculus Early Transcendentals

Question

Find the volume of the following solids. The solid bounded by the paraboloids \(z=2 x^{2}+y^{2}\) and \(z=27-x^{2}-2 y^{2}\)

Step-by-Step Solution

Answer

Question: Find the volume of the solid bounded by the paraboloids \(z = 2x^2 + y^2\) and \(z = 27 - x^2 - 2y^2\). Answer: The volume of the solid bounded by the two paraboloids is \(\frac{81}{2}\pi\).

Step 1: Identify equation for intersection curves

We first need to find where these two paraboloids intersect each other. Let's set the z-coordinates of the two equations equal to each other and solve for x and y: $$2x^2 + y^2 = 27 - x^2 - 2y^2$$ Now, we will rewrite this equation as a single equation in x and y coordinate, which represents the intersection curves of the paraboloids.

Step 2: Rewrite the equation

Let's rewrite the equation as follows: $$3x^2 + 3y^2 = 27$$ Divide the equation by 3: $$x^2 + y^2 = 9$$ The intersection curve is a circle with a center at the origin and radius of 3 in the xy-plane.

Step 3: Determine the volume bounds

Now, let's find the volume bounds. Our goal is to integrate over the region enclosed by these two surfaces along the z-axis. We can parameterize this region by cylindrical coordinates: $$x = r\cos(\theta)$$ $$y = r\sin(\theta)$$ Since the intersection curve is a circle of radius 3, the bounds for r and θ are: $$0 \leq r \leq 3$$ $$0 \leq \theta \leq 2\pi$$ For the z bounds, we will find the minimum and maximum z-coordinate of the solid at a given point (r, θ). We can express the two surfaces in cylindrical coordinates to find the minimum and maximum z values: 1) \(z = 2r^2\cos^2(\theta) + r^2\sin^2(\theta)\) 2) \(z = 27 - r^2\cos^2(\theta) - 2r^2\sin^2(\theta)\)

Step 4: Set up the triple integral

Now we need to set up the triple integral which will find the volume of the solid. We can write the integral with respect to dz, dr, and dθ in the following order: $$V = \int_{0}^{2\pi} \int_{0}^{3} \int_{2r^2\cos^2(\theta) + r^2\sin^2(\theta)}^{27 - r^2\cos^2(\theta) - 2r^2\sin^2(\theta)} rdzdrdθ$$

Step 5: Calculate the triple integral

Now we can calculate the triple integral: $$V = \int_{0}^{2\pi} \int_{0}^{3} [27 - r^2\cos^2(\theta) - 2r^2\sin^2(\theta) - (2r^2\cos^2(\theta) + r^2\sin^2(\theta))]r dr d\theta$$ Simplify the integrand: $$V = \int_{0}^{2\pi} \int_{0}^{3} (27 - 3r^2\cos^2(\theta) - 3r^2\sin^2(\theta))r dr d\theta$$ We can factor out the 3 from the integrand: $$V = \int_{0}^{2\pi} \int_{0}^{3} (27 - 3(r^2\cos^2(\theta) + r^2\sin^2(\theta)))r dr d\theta$$ Now, we can notice that \(\cos^2(\theta) + \sin^2(\theta) = 1\). So, the integrand simplifies to: $$V = \int_{0}^{2\pi} \int_{0}^{3} (27 - 3r^2)r dr d\theta$$ Now, let's calculate the integral: $$V = \int_{0}^{2\pi} \left[\int_{0}^{3} (27r - 3r^3) dr \right] d\theta$$ Integrate with respect to r: $$V = \int_{0}^{2\pi} \left[\frac{27}{2}r^2 - \frac{3}{4}r^4 \right]_{0}^{3} d\theta$$ Substitute r = 3 and subtract the result of r = 0: $$V = \int_{0}^{2\pi} [27(3)^2/2 - 3(3)^4/4] d\theta$$ Simplify: $$V = \int_{0}^{2\pi} \left[\frac{81}{2} - \frac{81}{4} \right] d\theta$$ Calculate the integral with respect to θ: $$V = \left[\frac{81}{4}\theta \right]_{0}^{2\pi}$$ Substitute θ = 2π and subtract the result of θ = 0: $$V = \frac{81}{4}(2\pi)$$ Thus, the volume of the solid bounded by the two paraboloids is: $$V = \frac{81}{2}\pi$$

Key Concepts

Triple Integrals

The concept of a triple integral is an extension of double integrals into three dimensions.
Triple integrals are incredibly useful when calculating the volume of a solid region in space.
By using a triple integral, you can find the cumulative 'sum' of infinitesimally small volumes, stacked up in three-dimensional space.

• Start by identifying the region of integration in your space.
• Apply the function you are integrating. This could represent anything from density to volume.
• Integrate successively over three variables, which usually represent x, y, and z.

For the problem at hand, triple integrals help us measure the solid region bounded by the intersecting paraboloids by integrating over z (the height) first, then over the radial and angular distance (r and \(\theta\)) from the cylindrical coordinates.

Cylindrical Coordinates

Cylindrical coordinates are an efficient way to handle problems that have circular symmetry, like the solid in our problem. They transform a three-dimensional Cartesian point (x, y, z) into a set of (r, \(\theta\), z) coordinates.

• \(r\) represents the distance from the z-axis to the point in the xy-plane.
• \(\theta\) is the angle formed with the x-axis.
• \(z\) remains the height.

This system is particularly useful for our problem because the intersection of two paraboloids is a circular region in the xy-plane. By using cylindrical coordinates, calculations become much simpler since the bounds of \(r\) and \(\theta\) directly correspond to the radius and circumference of the circle.

Paraboloids

Paraboloids are a type of quadric surface that can be either elliptic or hyperbolic. In our problem, we're dealing with elliptical paraboloids defined by their equations in a three-dimensional space.
The paraboloids in this problem have equations:

• \(z = 2x^2 + y^2\)
• \(z = 27 - x^2 - 2y^2\)

The first equation shows an upward opening paraboloid, while the second opens downward. These equations determine the shape and boundary of the solid whose volume we're trying to find. Understanding their equations helps us set up the limits for our triple integral accurately, as each paraboloid determines a range of permissible values for z for each point in the region defined.

Intersection Curves

Intersection curves arise when two surfaces meet, forming a new curve along their common boundary. For our problem, the paraboloids intersect along a circle in the xy-plane.

• To find it, equate the two equations \(2x^2 + y^2 = 27 - x^2 - 2y^2\).
• Simplifying gives \(x^2 + y^2 = 9\).

This equation describes a circle with a radius of 3, centered at the origin, lying flat in the xy-plane.This intersection provides one of the limits for our integral in cylindrical coordinates because it forms the boundary around which the region in space is enclosed. By understanding the intersection, we accurately limit the domain for our integration, ensuring we calculate the volume correctly.