Problem 58 - University Calculus: Early Transcendentals

Question

Find the volume of the region enclosed by the cylinder \(x^{2}+y^{2}=4\) and the planes \(z=0\) and \(x+y+z=4\).

Step-by-Step Solution

Answer

The volume is approximately 16 cubic units.

Step 1: Understand the Cylinder

The equation of the cylinder is given by \(x^2 + y^2 = 4\). This describes a circular cylinder with a radius of 2, oriented along the z-axis. Points on the cylinder satisfy that they are at a constant distance of 2 units from the z-axis.

Step 2: Understand the Planes

We have the plane \(z = 0\), which is the xy-plane, and another plane \(x + y + z = 4\). This second plane can be rewritten as \(z = 4 - x - y\). This equation implies that the plane is slanted, intercepting the z-axis at z = 4 when x and y are zero.

Step 3: Determine the Intersection

At the top, the cylinder and the slanted plane intersect when \(z = 4 - x - y\) across the boundary of the cylinder. Substitute \(x^2 + y^2 = 4\) into the plane's equation to find the bounds within the xy-plane.

Step 4: Set Up the Volume Integral

The volume can be found by integrating the plane’s height function over the circular region: \[\text{Volume} = \int_{-2}^{2} \int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}} (4 - x - y) \, dy \, dx\]This double integral calculates the volume from the circular base at \(z=0\) to the plane.

Step 5: Solve the Inner Integral

The inner integral is with respect to \(y\): \[\int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}} (4 - x - y) \, dy\]Breaking it down,\[= \int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}} 4 - x \, dy - \int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}} y \, dy \]Calculate these separately.

Step 6: Calculate the Inner Integrals

For the first here:\[= (4-x) \times 2\sqrt{4-x^2}\]For the second:\[\int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}} y \, dy = 0\]The terms involving \(y\) cancel out due to symmetry about \(y\), so its contribution is zero.

Step 7: Solve the Outer Integral

Substitute back into the outer integral and solve:\[\int_{-2}^2 2(4 - x) \sqrt{4-x^2} \, dx\]Here, use a trigonometric substitution or numeric computation for an exact or approximate solution to evaluate this definite integration, which completes the calculation.

Key Concepts

Volume of Solids

Understanding the volume of a solid enclosed by geometric shapes is a central topic in calculus. It's about determining the space an object occupies. To find the volume of a solid, especially with complex boundaries like cylinders and planes, calculus provides powerful techniques.
In this context, the volume refers to the 3-dimensional space within the boundaries we define. For the given exercise, we are focusing on a region enclosed by a cylinder and two planes. Calculating such volumes often involves setting up and evaluating integrals that respect these geometric constraints.

In this exercise, one plane is flat (the xy-plane at \(z = 0\)), and the other plane is inclined (\(x + y + z = 4\)), providing varying heights as \(x\) and \(y\) change. The challenge is to integrate the height of the volume under these conditions across a circular base defined by the cylinder (\(x^2 + y^2 = 4\)).

Double Integrals

Double integrals are a crucial concept when finding volumes under surfaces. They extend the idea of integration to functions of two variables, normally over some region in the xy-plane.
For this problem, we use a double integral to find the volume below the plane and above the base circle. The integral calculates the accumulated 'height' of the solid from the xy-plane up to the slanted plane.

The steps involve:

• Identifying the region of integration, here the disk defined by the cylinder base (\(x^2 + y^2 \leq 4\)).
• Setting up the integration limits using the circle's equation, which bounds \(y\) for any \(x\): from \(-\sqrt{4-x^2}\) to \(\sqrt{4-x^2}\).
• Solving inner integrals to simplify, followed by solving the outer integral.

The double integral reflects how calculus can calculate volumes by summing infinitely many infinitesimally thin layers of area.

Cylinders in Calculus

Cylinders are omnipresent in calculus problems involving volumes and surfaces. They provide a symmetrical, constant-radius surface, making it easier to integrate over.
When a cylinder like \(x^2 + y^2 = 4\) is presented, it's essential to understand its geometric implications. This cylinder is oriented along the z-axis with a radius of 2. All points on the cylinder are consistently 2 units from the z-axis, highlighting why the boundary curves as you move up or down.

In calculus, intersecting shapes like planes with cylinders create unique regions to examine. Here, the cylinder's mathematical description directly leads us to suitable limits of integration. The circle, described by its equation within the xy-plane, bounds the region over which we integrate to find volume.

Planes in Calculus

Planes in calculus are flat surfaces that simplify evaluating volumes and areas above and below them. In our example, we have two planes: one at \(z=0\), representing the base, and a slanted plane given by \(x + y + z = 4\).

For the plane \(z=0\):

• It serves as the base or bottom boundary of our region of interest.

For the plane \(x + y + z = 4\):

• It intercepts the z-axis at 4, tilts equally in directions of x and y, establishing a slant.
• This plane provides a changing height value in our volume calculation, depending on \(x\) and \(y\).

Understanding planes in the context of integration is important, as they define perspectives through which volumes and areas beneath them can be computed analytically or numerically.