Calculating Volume Enclosed By Paraboloids A Step-by-Step Guide

Hey guys! Ever wondered how to find the volume of a funky 3D shape formed by the intersection of two paraboloids? It might sound intimidating, but with a little calculus magic, we can totally nail it! Let's dive into how to calculate the volume of the solid enclosed by the paraboloids z = 4(x² + y²) and z = 2 - 4(x² + y²). We will break down the process step-by-step, making it super easy to follow.

Understanding the Paraboloids

Before we jump into calculations, let's get a visual sense of what we're dealing with. Our first paraboloid, z = 4(x² + y²), opens upwards, with its vertex at the origin (0, 0, 0). Think of it like a bowl sitting right-side up. The second paraboloid, z = 2 - 4(x² + y²), opens downwards and is centered along the z-axis. This one is like an upside-down bowl. The volume we are after is the space trapped between these two surfaces, where they intersect. To really grasp this, imagine the two bowls intersecting; the volume we want is the 3D region enclosed within their overlapping space.

Visualizing these paraboloids helps us understand the problem better. The upward-opening paraboloid z = 4(x² + y²) starts at the origin and expands upwards as x and y move away from zero. The downward-opening paraboloid z = 2 - 4(x² + y²) starts at z = 2 and opens downwards. The intersection of these two paraboloids will create a closed region in 3D space, and our goal is to find the volume of this region. We need to determine where these two surfaces intersect, as that will give us the bounds of our integration. Understanding the geometry of these shapes is crucial for setting up the integral correctly, ensuring we capture the entire volume enclosed between the surfaces.

Finding the Intersection

The first crucial step in finding the volume is determining where these paraboloids intersect. This intersection will define the boundaries of our solid. To find this, we set the equations equal to each other: 4(x² + y²) = 2 - 4(x² + y²). Adding 4(x² + y²) to both sides gives us 8(x² + y²) = 2. Dividing both sides by 8, we get x² + y² = 1/4. Ah-ha! This equation represents a circle in the xy-plane with a radius of 1/2. This circular region is the projection of the intersection onto the xy-plane and will be vital for setting up our double integral.

Understanding the intersection is paramount because it defines the limits of integration. The equation x² + y² = 1/4 tells us that the projection of the intersection onto the xy-plane is a circle centered at the origin with a radius of 1/2. This means that when we integrate, we will be integrating over this circular region. This step is critical for setting up the integral correctly and ensuring we calculate the volume accurately. The intersection not only gives us the bounds in the xy-plane but also helps us understand the shape of the solid we are trying to measure. Knowing the radius of the circle (1/2) allows us to define the limits for polar coordinates, which will simplify the integration process significantly.

Setting Up the Integral

Okay, now that we know the paraboloids intersect in a circle, we can set up a double integral to find the volume. The general idea is to integrate the difference between the two surfaces over the region of intersection. Think of it as stacking infinitesimally thin layers on top of each other within the enclosed space. The volume element in Cartesian coordinates is dV = dz dy dx, but since we have a circular region, polar coordinates will be our best friend here! Let's switch to polar: x = r cos θ, y = r sin θ, and x² + y² = r². The Jacobian for this transformation is r, so our volume element becomes dV = r dz dr dθ. The beauty of polar coordinates is that they perfectly match the circular symmetry of our problem, making the integral much easier to solve.

In polar coordinates, the equation x² + y² = 1/4 becomes r² = 1/4, so r ranges from 0 to 1/2. The angle θ will cover the entire circle, so it ranges from 0 to 2π. Our paraboloid equations become z = 4r² and z = 2 - 4r². The limits of integration for z will be from the lower surface (z = 4r²) to the upper surface (z = 2 - 4r²). Thus, our triple integral transforms into a more manageable double integral. Setting up the integral correctly is essential for obtaining the correct volume. We are essentially summing up the volume of infinitesimal cylinders over the circular region. The integrand represents the height of these cylinders (the difference in z-values), and the limits of integration define the region over which we are summing these volumes. This methodical approach ensures we capture the entire volume of the solid.

Evaluating the Integral

Alright, time for some calculus action! Our integral is set up, and now we need to evaluate it. The integral we need to compute is: ∫₀²π ∫₀¹/₂ (2 - 4r² - 4r²) r dr dθ. Let's simplify the integrand first: (2 - 8r²)r = 2r - 8r³. Now we have ∫₀²π ∫₀¹/₂ (2r - 8r³) dr dθ. First, we integrate with respect to r: ∫ (2r - 8r³) dr = r² - 2r⁴. Evaluating this from 0 to 1/2, we get (1/2)² - 2(1/2)⁴ = 1/4 - 2(1/16) = 1/4 - 1/8 = 1/8. So our integral simplifies to ∫₀²π (1/8) dθ. This is a simple integral with respect to θ, and we get (1/8)θ evaluated from 0 to 2π, which is (1/8)(2π) = π/4. And there we have it! The volume of the solid enclosed by the two paraboloids is π/4 cubic units.

Breaking down the integral into steps makes it much easier to manage. We first simplified the integrand by combining like terms and then integrated with respect to r. After evaluating the inner integral, we were left with a simple integral with respect to θ. This methodical approach ensures we make no mistakes in the integration process. The final result, π/4, gives us the volume of the solid enclosed by the two paraboloids. This calculation highlights the power of calculus in solving complex geometric problems. By setting up the integral correctly and using appropriate coordinate systems, we can find the volumes of even the most complicated shapes.

Alternative approach using triple integrals

We can calculate the volume using a triple integral directly in cylindrical coordinates. The setup involves integrating over the region enclosed by the two paraboloids, z = 4(x² + y²) and z = 2 - 4(x² + y²). The cylindrical coordinates (r, θ, z) are defined as x = r cos θ, y = r sin θ, and z = z. The Jacobian for the transformation from Cartesian to cylindrical coordinates is r, so dV = r dz dr dθ. The advantage of using cylindrical coordinates is that the equations of the paraboloids become simpler: z = 4r² and z = 2 - 4r².

Setting up the Triple Integral

To set up the triple integral, we first need to determine the limits of integration. As we found earlier, the intersection of the two paraboloids projects onto the xy-plane as a circle with the equation x² + y² = 1/4, which in cylindrical coordinates is r² = 1/4 or r = 1/2. Thus, r ranges from 0 to 1/2. The angle θ ranges from 0 to 2π to cover the entire circle. The z limits are determined by the paraboloids themselves, ranging from the lower surface z = 4r² to the upper surface z = 2 - 4r². The triple integral for the volume V is given by:

V = ∫₀²π ∫₀¹/₂ ∫₄r²²⁻⁴r² r dz dr dθ

This integral represents the sum of infinitesimally small volume elements r dz dr dθ over the region enclosed by the paraboloids. The innermost integral integrates with respect to z, representing the height of a cylindrical shell. The middle integral integrates with respect to r, representing the radius of the shell, and the outermost integral integrates with respect to θ, covering the entire circle.

Evaluating the Triple Integral

Now, let's evaluate the triple integral step by step. First, we integrate with respect to z:

∫₄r²²⁻⁴r² r dz = r[z]₄r²²⁻⁴r² = r(2 - 4r² - 4r²) = r(2 - 8r²)

Next, we integrate with respect to r:

∫₀¹/₂ r(2 - 8r²) dr = ∫₀¹/₂ (2r - 8r³) dr = [r² - 2r⁴]₀¹/₂ = (1/4 - 2(1/16)) - (0) = 1/4 - 1/8 = 1/8

Finally, we integrate with respect to θ:

∫₀²π (1/8) dθ = (1/8)[θ]₀²π = (1/8)(2π) = π/4

So, the volume of the solid enclosed by the two paraboloids is π/4 cubic units, which matches the result we obtained using the double integral method. This confirms the correctness of our calculations and demonstrates the versatility of triple integrals in finding volumes of 3D solids.

Comparing the Double and Triple Integral Methods

Both the double and triple integral methods provide the same result, π/4 cubic units, for the volume enclosed by the paraboloids. The double integral method involves integrating the difference between the two surfaces over the projection of their intersection onto the xy-plane. This approach is often simpler when the region of integration in the xy-plane is easily described, as in this case with the circle x² + y² = 1/4.

The triple integral method, on the other hand, directly integrates over the 3D region enclosed by the surfaces. It can be more straightforward to set up when the boundaries of the region are naturally expressed in three dimensions. In this case, using cylindrical coordinates simplified the integral setup by aligning with the symmetry of the paraboloids.

The choice between the two methods often depends on the specific problem and personal preference. For problems with clear projections onto a plane, the double integral method may be more efficient. For problems with complex 3D boundaries, the triple integral method may provide a more direct approach. Both methods, however, rely on a solid understanding of multivariable calculus and the geometry of the surfaces involved.

Conclusion

So, there you have it! We've successfully found the volume of the solid trapped between those two paraboloids. Whether you choose to use double or triple integrals, the key is to break the problem down into manageable steps: visualize the surfaces, find their intersection, set up the integral in the most convenient coordinate system, and evaluate. With practice, these kinds of volume calculations become much easier. Keep exploring those 3D shapes, and happy calculating!