Problem 37 - Thomas Calculus

Question

Find the area of the surface cut from the paraboloid \(x^{2}+y^{2}-z=\) 0 by the plane \(z=2\) .

Step-by-Step Solution

Answer

The area is \(\frac{53\pi}{6}\).

Step 1: Identify the Intersection Curve

The given surface is a paraboloid described by the equation \(x^2 + y^2 = z\). The plane \(z = 2\) cuts through this paraboloid. First, we substitute \(z = 2\) into the paraboloid equation to find the intersection curve: \(x^2 + y^2 = 2\). This equation represents a circle in the \(xy\)-plane with radius \(\sqrt{2}\).

Step 2: Set Up the Surface Area Integral

The area of the surface cut from the paraboloid by the plane corresponds to the surface area of the paraboloid above the circle \(x^2 + y^2 = 2\). The surface described by \(z = x^2 + y^2\) can have its area calculated in polar coordinates. The parametric representation of the surface leads to \(\mathbf{r}(r, \theta) = (r \cos \theta, r \sin \theta, r^2)\), with polar limits \(0 \leq r \leq \sqrt{2}\) and \(0 \leq \theta < 2\pi\).

Step 3: Calculate the Partial Derivatives

We calculate the partial derivatives \(\frac{\partial \mathbf{r}}{\partial r}\) and \(\frac{\partial \mathbf{r}}{\partial \theta}\). These derivatives are: \(\frac{\partial \mathbf{r}}{\partial r} = (\cos \theta, \sin \theta, 2r)\) and \(\frac{\partial \mathbf{r}}{\partial \theta} = (-r \sin \theta, r \cos \theta, 0)\).

Step 4: Compute the Cross Product

Compute the cross product \(\frac{\partial \mathbf{r}}{\partial r} \times \frac{\partial \mathbf{r}}{\partial \theta}\). This results in: \[ (\cos \theta, \sin \theta, 2r) \times (-r \sin \theta, r \cos \theta, 0) = (-2r^2 \cos \theta, -2r^2 \sin \theta, r) \]

Step 5: Determine the Magnitude

Find the magnitude of the cross product vector to compute the differential area: \[ \sqrt{(-2r^2 \cos \theta)^2 + (-2r^2 \sin \theta)^2 + r^2} = \sqrt{4r^4 + r^2} = r\sqrt{4r^2 + 1} \]

Step 6: Integrate to Find Total Surface Area

Integrate the expression for surface area over the region defined by the intersection. The surface area \(A\) is given by:\[ A = \int_0^{2\pi} \int_0^{\sqrt{2}} r\sqrt{4r^2 + 1} \, dr \, d\theta \]First compute the integral with respect to \(r\), then integrate with respect to \(\theta\).

Step 7: Compute the Inner Integral

Calculate the integral \(\int_0^{\sqrt{2}} r\sqrt{4r^2 + 1} \, dr\). Make the substitution \(u = 4r^2 + 1\), \(du = 8r \, dr\), leading to:\[ \frac{1}{8} \int_{1}^{9} \sqrt{u} \, du = \frac{1}{8} \times \left[ \frac{2}{3} u^{3/2} \right]_{1}^{9} = \frac{1}{12} [9\sqrt{9} - 1]\]This simplifies to \( \frac{1}{12} (54 - 1) = \frac{53}{12} \).

Step 8: Evaluate the Final Integral

Substitute the inner integral result into the outer integral to find the full area:\[ A = \int_0^{2\pi} \frac{53}{12} \, d\theta = \frac{53}{12} \times [\theta]_0^{2\pi} = \frac{53}{12} \times 2\pi = \frac{53\pi}{6} \]

Step 9: Final Result

Thus, the total area of the surface cut from the paraboloid \(x^2 + y^2 - z = 0\) by the plane \(z=2\) is \(\frac{53\pi}{6}\).

Key Concepts

Intersection of Surfaces

When dealing with three-dimensional figures, understanding how they intersect is crucial. Here, we are examining the intersection between a paraboloid and a plane. The paraboloid is given by the equation \( x^2 + y^2 = z \), while the plane is represented by \( z = 2 \). Replacing \( z \) in the paraboloid equation with 2 gives us \( x^2 + y^2 = 2 \). This equation describes a circle with radius \( \sqrt{2} \). This circle represents the path along which the plane slices through the paraboloid. Grasping this intersection helps in visualizing the surface area trapped "above" this circle, as we further calculate.

Polar Coordinates

Polar coordinates offer a simplified method to handle calculations involving circles and curves. Instead of relying on rectangular coordinates (\(x, y\)), polar coordinates use a distance \(r\) from the origin and an angle \(\theta\) from the polar axis. Here, we use polar coordinates to represent every point on the intersection circle \( x^2 + y^2 = 2 \). The switch to polar coordinates,

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

makes it convenient to integrate circular regions. For this integration, \( r \) ranges from 0 to \( \sqrt{2} \), and \( \theta \) spans from \( 0 \) to \( 2\pi \). This setup aligns perfectly with the form of the paraboloid as we examine the surface contained within the bounding circle.

Parametric Representation

A parametric representation is a method of expressing a surface with parameters \( r \) and \( \theta \). This approach is particularly useful for integrating over surfaces in three dimensions. For the paraboloid \( z = x^2 + y^2 \), the parametric representation becomes \( \mathbf{r}(r, \theta) = (r \cos \theta, r \sin \theta, r^2) \). This formula maps the surface in terms of \( r \) and \( \theta \), helping us visualize and compute properties like surface area. As we change \( r \) and \( \theta \), this parametric form allows us to accurately describe the 3D surface, essentially "drawing" the paraboloid section using geometry and calculus.

Integral Calculus

Integral calculus is an essential tool in finding the area of complex surfaces. In this problem, we integrate over the parametric representation of the paraboloid described using polar coordinates. After setting up the bounds and the parametric form, the focus moves to finding the area by evaluating an integral. We compute the partial derivatives of the parametric surface, apply a cross product, and determine its magnitude. This magnitude, representing differential area elements, forms the integral's core component:

• Cross Product: Computes surface orientation
• Magnitude: Represents the differential area

The final integral, \[ A = \int_0^{2\pi} \int_0^{\sqrt{2}} r\sqrt{4r^2 + 1} \, dr \, d\theta \], evaluates the surface area by integrating first with respect to \( r \) and subsequently \( \theta \). This process simplifies the problem, turning a theoretical surface problem into computations using algebra and calculus principles.