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Magazine Chapter 15: Problem 83
Cylinder and sphere Find the volume of the region cut from the solid cylinder \(x^{2}+y^{2} \leq 1\) by the sphere \(x^{2}+y^{2}+z^{2}=4\)
Short Answer
Expert verified
The volume of the intersection is \(2\pi\sqrt{3}\).
Step by step solution
01
Define the Region of Intersection
The solid cylinder is described by the inequality \(x^2 + y^2 \leq 1\), which spans infinite heights along the \(z\)-axis. The sphere is given by \(x^2 + y^2 + z^2 = 4\). The region of intersection between the cylinder and the sphere must satisfy both equations.
02
Find Bounds for z
Substitute \(x^2 + y^2 = 1\) (the boundary of the cylinder) into the sphere's equation: \(x^2 + y^2 + z^2 = 4\), which simplifies to \(z^2 = 3\). Therefore, the bounds for \(z\) are \(- \sqrt{3} \leq z \leq \sqrt{3}\). This describes how the cylinder intersects vertically with the sphere.
03
Set Up the Triple Integral for Volume
The volume of the region can be represented as \[ V = \int_{-\sqrt{3}}^{\sqrt{3}} \int_{-1}^{1} \int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}} dx \, dy \, dz \]. The integral accounts for the cylindrical bounds in the \(x\) and \(y\) directions and the intersection bounds in the \(z\) direction.
04
Integrate with respect to x
First, integrate over \(x\): \[ \int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}} dx = 2\sqrt{1-y^2} \]. This represents the horizontal cross-section of the cylinder at a given \(y\).
05
Integrate with respect to y
Next, integrate over \(y\): \[\int_{-1}^{1} 2\sqrt{1-y^2} \, dy\]. This integral is the area of a semicircle with radius 1, which results in \(\pi\).
06
Integrate with respect to z
Finally, integrate over \(z\): \[\int_{-\sqrt{3}}^{\sqrt{3}} \pi \, dz = \pi \cdot (2\sqrt{3}) = 2\pi\sqrt{3}\]. This calculates the volume of the region cut from the cylinder by the sphere.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Cylinder
A cylinder is a three-dimensional geometric shape with two parallel circular bases connected by a curved surface. In calculus, the cylinder is often described by inequalities such as
This results in infinitely extending along the \(z\)-axis unless bounded by another geometric shape, such as a sphere or a plane. The projection of a cylinder onto the xy-plane is a circle with radius 1.
- The cylindrical boundary: \[ x^2 + y^2 \leq 1 \].
- The vertical extent is along the \(z\)-axis, which in this problem is not limited.
This results in infinitely extending along the \(z\)-axis unless bounded by another geometric shape, such as a sphere or a plane. The projection of a cylinder onto the xy-plane is a circle with radius 1.
Sphere
A sphere is defined in three-dimensional space by the set of all points that are equidistant from a given point, called the center. In this exercise, the sphere is described by the equation:
\[ x^2 + y^2 + z^2 = 4 \].This equation specifies a sphere with a radius of 2 units, centered at the origin \((0,0,0)\). The sphere encompasses a spherical volume, cutting through space.
\[ x^2 + y^2 + z^2 = 4 \].This equation specifies a sphere with a radius of 2 units, centered at the origin \((0,0,0)\). The sphere encompasses a spherical volume, cutting through space.
- When intersecting other shapes, such as cylinders, the resulting region is often a complex shape.
- The boundary of this sphere is where its equation equals exactly 4, but within this boundary, any point satisfies the inequality \(x^2 + y^2 + z^2 \leq 4\).
Triple Integral
Triple integrals extend the concept of integration to three dimensions, which allows us to calculate volumes of unusual shapes. A triple integral is generally expressed as:
\[ \int \int \int f(x,y,z) \, dx \, dy \, dz \]Where \(f(x,y,z)\) is the function being integrated over a given volume. In our case:
\[ \int \int \int f(x,y,z) \, dx \, dy \, dz \]Where \(f(x,y,z)\) is the function being integrated over a given volume. In our case:
- The integrand is simply \(1\) because we are calculating a volume.
- To find the volume of the region intersected by the cylinder and sphere, we use a triple integral bounded by the respective constraints of these shapes.
- The integration limits first differ along the \(x\), \(y\), and \(z\) coordinates.
Volume Calculation
Volume calculation using a triple integral involves evaluating the integral step-by-step. Each integral simplifies our expression:
- The first step is to integrate over \(x\), resulting in the expression \(2\sqrt{1-y^2}\), which is the width at any slice \(y\).
- Next, integrating over \(y\) involves finding the area under the semicircle defined by \(\sqrt{1-y^2}\), which equals \(\pi\).
- Finally, integrating over \(z\), with bounds \(-\sqrt{3}\) to \(\sqrt{3}\), involves multiplying the area result by the height of the intersection, yielding the total volume.
