Calculating Volume Sphere, XY-Plane, And Cone A Multivariable Calculus Problem

Hey guys! Today, we're diving into a super cool problem in multivariable calculus – finding the volume of a solid that's defined by some pretty interesting boundaries. We're talking about a solid that lives inside a sphere, floats above the xy-plane, and manages to stay outside a cone. Sounds like a geometrical party, right? Let's break it down step by step.

Problem Statement

So, here's the deal. We need to find the volume of the solid that satisfies these three conditions:

1. It lies within the sphere x^2 + y^2 + z^2 = 49.
2. It's above the xy-plane (meaning z ≥ 0).
3. It's outside the cone z = 7√(x^2 + y^2).

This is a classic problem that combines our understanding of spheres, cones, and how to use triple integrals to calculate volumes. Buckle up, because we're about to embark on a mathematical adventure!

Visualizing the Solid

Before we start crunching numbers, it's super helpful to get a mental picture of what we're dealing with. Imagine a sphere centered at the origin with a radius of 7 (since 49 is 7 squared). Now, picture a cone that opens upwards, also with its vertex at the origin. The equation z = 7√(x^2 + y^2) tells us that this is a pretty steep cone.

We're only interested in the part of the sphere that's above the xy-plane, so we can think of it as a hemisphere. Then, we carve out a chunk of this hemisphere using the cone. The solid we want is the stuff that's left – the volume inside the hemisphere but outside the cone. Visualizing this 3D shape is key to setting up our integral correctly. Think of it like scooping out ice cream from a hemispherical bowl with a cone-shaped scoop – what's left is what we need to measure!

Choosing the Right Coordinate System

Okay, now that we have a good picture in our minds, it's time to think about how we're going to calculate this volume. We could try to do it using rectangular coordinates (x, y, z), but trust me, that would be a mess. The equations for the sphere and the cone would make the limits of integration super complicated.

The smart move here is to switch to spherical coordinates. Spherical coordinates are like the GPS for 3D space, using three parameters: ρ (rho), φ (phi), and θ (theta).

• ρ (rho) is the distance from the origin to the point, kind of like the radius in polar coordinates but in 3D.
• φ (phi) is the angle from the positive z-axis, ranging from 0 to π.
• θ (theta) is the same angle as in polar coordinates, measuring the angle around the z-axis from 0 to 2π.

The beauty of spherical coordinates is that they play really nicely with spheres and cones. Our sphere's equation, x^2 + y^2 + z^2 = 49, becomes simply ρ^2 = 49, or ρ = 7. That's way easier to deal with!

Transforming to Spherical Coordinates

Before we can set up our integral, we need to express our other equations in spherical coordinates too. Remember these key relationships:

• x = ρ sin φ cos θ
• y = ρ sin φ sin θ
• z = ρ cos φ

Let's tackle the cone equation, z = 7√(x^2 + y^2). We can substitute our spherical coordinate equivalents:

ρ cos φ = 7√(ρ^2 sin^2 φ cos^2 θ + ρ^2 sin^2 φ sin^2 θ)

Notice how we can factor out ρ^2 sin^2 φ from under the square root:

ρ cos φ = 7ρ sin φ √(cos^2 θ + sin^2 θ)

Since cos^2 θ + sin^2 θ = 1, we get:

ρ cos φ = 7ρ sin φ

We can divide both sides by ρ (assuming ρ isn't zero, which it won't be in our region) and get:

cos φ = 7 sin φ

Now, divide both sides by cos φ:

1 = 7 tan φ

So, tan φ = 1/7. This means φ = arctan(1/7). Let's call this angle φ₀. This angle represents the cone in spherical coordinates. Everything outside the cone will have an angle φ less than φ₀.

Setting Up the Triple Integral

Alright, we're in the home stretch! We now have everything we need to set up our triple integral in spherical coordinates. Remember, the volume element in spherical coordinates is:

dV = ρ^2 sin φ dρ dφ dθ

This is a super important factor that accounts for the stretching and squishing that happens when we transform from rectangular to spherical coordinates. Don't forget it!

Now, let's think about our limits of integration:

• ρ (rho): The distance from the origin ranges from 0 (the origin) to 7 (the sphere's radius).
• φ (phi): The angle from the z-axis ranges from 0 (the z-axis) to arctan(1/7) (the cone). Since we're above the xy-plane, we stop at π/2.
• θ (theta): The angle around the z-axis ranges from 0 to 2π (a full circle).

So, our triple integral for the volume looks like this:

V = ∫₀^(2π) ∫₀^(arctan(1/7)) ∫₀⁷ ρ^2 sin φ dρ dφ dθ

Evaluating the Integral

Time to put our calculus skills to work! Let's evaluate this triple integral step by step.

First, we integrate with respect to ρ:

∫₀⁷ ρ^2 sin φ dρ = sin φ [ρ³/3]₀⁷ = (343/3) sin φ

Now, we integrate with respect to φ:

∫₀^(arctan(1/7)) (343/3) sin φ dφ = (343/3) [-cos φ]₀^(arctan(1/7))

This is where things get a little trigonometric. We need to find cos(arctan(1/7)). Remember that arctan(1/7) gives us an angle whose tangent is 1/7. We can think of this as a right triangle with opposite side 1 and adjacent side 7. The hypotenuse would be √(1^2 + 7^2) = √50 = 5√2. So, cos(arctan(1/7)) = adjacent/hypotenuse = 7/(5√2).

Plugging this in, we get:

(343/3) [-7/(5√2) - (-1)] = (343/3) [1 - 7/(5√2)]

Finally, we integrate with respect to θ:

∫₀^(2π) (343/3) [1 - 7/(5√2)] dθ = (343/3) [1 - 7/(5√2)] [θ]₀^(2π) = (686π/3) [1 - 7/(5√2)]

So, the volume of the solid is (686π/3) [1 - 7/(5√2)] cubic units. We can simplify this a bit further if we want, but this is the exact answer.

Final Thoughts

Guys, we did it! We successfully navigated a challenging volume problem using spherical coordinates. This problem is a fantastic example of how choosing the right coordinate system can make a huge difference in the complexity of a calculation. Spherical coordinates are your best friend when dealing with spheres and cones!

Remember, the key to these problems is to visualize the solid, choose the appropriate coordinate system, set up the integral carefully, and then take your time evaluating it. Keep practicing, and you'll become a master of multivariable calculus in no time!