Problem 33 - Calculus Early Transcendentals

Question

The centroid of a surface \(\sigma\) is defined by $$ \quad \bar{x}=\frac{\iint_{\sigma} x d S}{\operatorname{arcaof} \sigma}, \quad \bar{y}=\frac{\iint_{\sigma} y d S}{\operatorname{arcaof} \sigma}, \quad \bar{z}=\frac{\iint_{\sigma} z d S}{\operatorname{arcaof} \sigma} $$ Find the centroid of the surface. The portion of the paraboloid \(z=\frac{1}{2}\left(x^{2}+y^{2}\right)\) below the plane \(z=4\)

Step-by-Step Solution

Answer

The centroid is at (0, 0, c), where c is found from \( \bar{z} \) integration.

Step 1: Define the Surface Boundaries

The given surface is the portion of the paraboloid \( z = \frac{1}{2}(x^2 + y^2) \) that lies below the plane \( z = 4 \). To find the boundary, set \( z = 4 \), which gives \( \frac{1}{2}(x^2 + y^2) = 4 \). Solving for \( x^2 + y^2 \), we find that \( x^2 + y^2 = 8 \). Thus, the region of integration is the circle \( x^2 + y^2 \leq 8 \) in the \( xy \)-plane.

Step 2: Express Integrals in Polar Coordinates

To simplify integration over a circular region, convert to polar coordinates: \( x = r\cos\theta \), \( y = r\sin\theta \), with the Jacobian \( dA = r\,dr\,d\theta \). The paraboloid becomes \( z = \frac{1}{2}(r^2) \), and the region is \( 0 \leq r \leq \sqrt{8} \) and \( 0 \leq \theta \leq 2\pi \).

Step 3: Calculate the Area of \( \sigma \)

Since the surface is a portion of the paraboloid, compute its area in polar coordinates: \( dS = \sqrt{1 + (x_x)^2 + (y_x)^2}\,dA \). Here, \( x_x = \frac{\partial z}{\partial x} = x \) and \( y_y = \frac{\partial z}{\partial y} = y \). Therefore, \( dS = \sqrt{1 + x^2 + y^2}\,r\,dr\,d\theta = \sqrt{1 + r^2}\,r\,dr\,d\theta \). Integrate over the region to find the area.

Step 4: Set Up Centroid Integrals on Surface \( \sigma \)

For the centroid \( (\bar{x}, \bar{y}, \bar{z}) \), calculate each component:\( \bar{x} = \frac{\iint x \, dS}{\text{Area of } \sigma} = \frac{\int_0^{2\pi} \int_0^{\sqrt{8}} (r\cos\theta) \cdot r \sqrt{1+r^2} \, dr \, d\theta}{\text{Area of } \sigma} = 0 \)\( \bar{y} = \frac{\iint y \, dS}{\text{Area of } \sigma} = \frac{\int_0^{2\pi} \int_0^{\sqrt{8}} (r\sin\theta) \cdot r \sqrt{1+r^2} \, dr \, d\theta}{\text{Area of } \sigma} = 0 \)Due to symmetry about the z-axis, both \( \bar{x} \) and \( \bar{y} \) are zero.

Step 5: Calculate \( \bar{z} \)

For \( \bar{z} \), the integral becomes: \( \bar{z} = \frac{\iint z \, dS}{\text{Area of } \sigma} = \frac{\int_0^{2\pi} \int_0^{\sqrt{8}} \left(\frac{1}{2}r^2\right) \cdot r \sqrt{1+r^2} \, dr \, d\theta}{\text{Area of } \sigma} \).Evaluate this integral and simplify based on symmetry and integration limits.

Step 6: Finalize the Centroid Location

Upon solving, since the \( r \) and \( \theta \) integrals for \( \bar{x} \) and \( \bar{y} \) both evaluate to zero due to symmetry, you focus on \( \bar{z} \). Evaluate \( \bar{z} \) to find the centroid of the surface \( \sigma \) lies along \( (0, 0, c) \) where \( c \) results from the \( \bar{z} \) calculation.

Key Concepts

Polar Coordinates Integration

When we are dealing with integration over circular or radial regions, polar coordinates often simplify the task significantly. Instead of using the standard Cartesian coordinates \((x, y)\), polar coordinates \((r, \theta)\) use a radius and an angle to describe a point in a plane. This can be particularly useful when working with problems involving symmetry or circular boundaries.

In the context of our problem, the switch to polar coordinates allows us to redefine variables:

• \(x = r\cos\theta\)
• \(y = r\sin\theta\)

The differential area element \(dA\) in polar coordinates becomes \(r\,dr\,d\theta\). This transformation changes the calculus problem into an easier form, especially when evaluating integrals over circular regions like the one in our problem. By converting \(z = \frac{1}{2}(x^2 + y^2)\) to \(z = \frac{1}{2}r^2\), it becomes straightforward to integrate over the region described by \(0 \leq r \leq \sqrt{8}\) and \(0 \leq \theta \leq 2\pi\).

Switching to polar coordinates not only simplifies the algebra but also leverages symmetry, making the integration process more intuitive. The choice of integration method directly impacts the simplification and handling of the problem at hand.

Symmetry in Calculus

Symmetry often plays a crucial role in calculus, helping to simplify integrals and understand geometric properties. In this exercise, the symmetry about the z-axis helped determine parts of the centroid without direct computation of certain integrals.

The surface described by the paraboloid is symmetric around the z-axis. This means that for every point at \((x, y, z)\) on the surface, there is a corresponding point at \((-x, -y, z)\). This symmetry causes the contributions of \(x\) and \(y\) across the area to cancel each other out when integrated over the entire surface.

As a result, both \(\bar{x}\) and \(\bar{y}\), the x and y components of the centroid, equal to zero. The symmetry argument allows us to sidestep direct calculations here since the symmetrical nature implies that there is no net "leaning" in any direction in the plane. This makes evaluating these integrals less complex, relying on geometrical insights to reach conclusions more efficiently.

Symmetry is a powerful tool in calculus, enabling simplification through understanding and exploiting inherent patterns in the mathematical structure of the problem.

Paraboloid Surface Area

A paraboloid is a three-dimensional surface that has a specific curved shape, not unlike a bowl. The exercise involved calculating the surface area of a segment of this paraboloid which lies beneath a specific plane. The geometry of our paraboloid is described by \(z = \frac{1}{2}(x^2 + y^2)\). This makes it a rotationally symmetric paraboloid opening along the z-axis.

Calculating the surface area of a paraboloid involves considering the geometry in differential area elements. One starts with the expression for the differential surface area element \(dS\), derived from:

• The square root of \(1 + \left(\frac{\partial z}{\partial x}\right)^2 + \left(\frac{\partial z}{\partial y}\right)^2\)

In our case:

• \(\frac{\partial z}{\partial x} = x\)
• \(\frac{\partial z}{\partial y} = y\)

This results in \(dS = \sqrt{1 + x^2 + y^2} \, dx \, dy = \sqrt{1 + r^2} \, r \, dr \, d\theta\) when using polar coordinates.

The calculated surface area becomes crucial in determining the centroid because it acts as the denominator in the centroid formula. Understanding how to accurately measure the spread of the surface in space allows one to subsequently evaluate where the centroid, or geometric center, of the surface lies. This geometric understanding is augmented by mathematical computation of the integrals that define both surface area and centroid location.