Problem 27 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the surface area of the given surface. If an integral cannot be evaluated using the Second Fundamental Theorem of Calculus, then use the Parabolic Rule with \(n=10 .\) The paraboloid \(z=x^{2}+y^{2}\) over the region (a) in the first quadrant and inside the circle \(x^{2}+y^{2}=9\) (b) inside the triangle with vertices \((0,0),(3,0),(0,3)\)

Step-by-Step Solution

Answer

Use polar coordinates for part (a) and integrate in Cartesian for part (b). Evaluate numerically if necessary.

Step 1: Understanding the Surface

The surface is defined by the equation of the paraboloid, \(z = x^2 + y^2\). We are given two regions to evaluate: one within a circle and one within a triangle.

Step 2: Setting Up for Part (a)

For part (a), the region is the first quadrant part of the circle \(x^2 + y^2 = 9\). We use polar coordinates to convert the region:\[\begin{align*} x &= r\cos\theta, \ y &= r\sin\theta, \ dx \, dy &= r \, dr \, d\theta. \end{align*}\] The limits are \(0 \leq \theta \leq \frac{\pi}{2}\) (first quadrant) and \(0 \leq r \leq 3\) (radius of the circle).

Step 3: Surface Area Integral for Part (a)

The formula for the surface area \(S\) of a parametrized surface \(z = f(x,y)\) is \[ S = \iint_R \sqrt{1 + \left(\frac{\partial z}{\partial x}\right)^2 + \left(\frac{\partial z}{\partial y}\right)^2} \, dx \, dy, \] where \(\frac{\partial z}{\partial x} = 2x\) and \(\frac{\partial z}{\partial y} = 2y\). Thus, \(\sqrt{1 + (2x)^2 + (2y)^2} = \sqrt{1 + 4(x^2 + y^2)}\). In polar coordinates, the integral becomes: \[ \int_0^{\frac{\pi}{2}} \int_0^3 \sqrt{1 + 4r^2} \, r \, dr \, d\theta. \]

Step 4: Evaluating the Integral for Part (a)

To evaluate \( \int_0^3 \sqrt{1 + 4r^2} \, r \, dr \), use substitution or numerical methods like the Parabolic Rule if needed. The result of this integral represents the surface area portion in the first quadrant.

Step 5: Setting Up for Part (b)

For part (b), the region is a triangle with vertices at \((0,0)\), \((3,0)\), and \((0,3)\). We use Cartesian coordinates here. The limits for \(x\) are \(0 \leq x \leq 3\), and for \(y\) are \(0 \leq y \leq 3-x\).

Step 6: Surface Area Integral for Part (b)

Using the surface area formula \[ S = \iint_R \sqrt{1 + 4(x^2 + y^2)} \, dx \, dy, \] we set up the integral with the limits derived in Step 5: \[ \int_0^3 \int_0^{3-x} \sqrt{1 + 4(x^2 + y^2)} \, dy \, dx. \]

Step 7: Evaluating the Integral for Part (b)

Evaluate \( \int_0^{3-x} \sqrt{1 + 4(x^2 + y^2)} \, dy \) for each \(x\) using a numerical method if necessary, then integrate with respect to \(x\). This gives the surface area over the triangular region.

Key Concepts

Paraboloid

A paraboloid is a three-dimensional surface shaped like an open bowl. Mathematically, it is described by the equation \( z = x^2 + y^2 \) in the given exercise. This is a standard form of a paraboloid known as an elliptic paraboloid, where the cross-sections parallel to the \(xy\)-plane are ellipses or circles, and those parallel to the \(xz\) or \(yz\) planes are parabolas.
It’s important to note that every point on the paraboloid has a coordinate \( (x, y, z) \), and the \(z\)-value is the sum of the squares of the \(x\) and \(y\) coordinates. Such a surface is involved when calculating geometric properties like surface area. By understanding the equation of a paraboloid, we can better comprehend how the surface extends in space, which is crucial when evaluating its surface area over a specified region.

Polar Coordinates

Polar coordinates are particularly useful for handling problems involving circular regions, as they simplify the and conversion of Cartesian coordinates into a radial coordinate system. Instead of using \((x, y)\), polar coordinates use \((r, \theta)\), where \(r\) is the distance from the origin, and \(\theta\) is the angle from the positive \(x\)-axis.
To convert Cartesian equations like those given in the problem, we use:

• \(x = r\cos\theta\)
• \(y = r\sin\theta\)
• \(dx \, dy = r \, dr \, d\theta\)

For example, in the exercise's Part (a), by switching from Cartesian to polar coordinates, the calculation of the surface area over the region constrained by \(x^2 + y^2 = 9\) becomes straightforward. Polar coordinates thus streamline such integrations by aligning the limits more naturally with circular or radial boundaries.

Numerical Integration

Numerical integration is a method used to approximate the value of an integral when a closed-form expression is difficult or impossible to find. This is particularly useful in cases where the integral is complex or involves functions that are hard to integrate analytically.
In our problem, numerical integration might be employed as specified when the integral cannot be resolved via the Second Fundamental Theorem of Calculus. One such method is the Parabolic Rule, also known as Simpson's Rule, which approximates the integral by fitting parabolas through segments of the function. This method improves accuracy by considering the curvature of the function across each segment, making it a reliable technique when dealing with complex surfaces like paraboloids.

Parametrized Surface

A parametrized surface is one in which each point on the surface can be described using parameters, typically denoted as \((u, v)\). For the problem at hand, we're looking at a surface described by the equation \(z = x^2 + y^2\), calculated over a specific region.
The parametrization allows us to express surface area through a formula involving partial derivatives, creating a basis for calculating how the surface stretches or shrinks over its span. In mathematical terms, the surface area \(S\) can be found using the integral:\[ S = \iint_R \sqrt{1 + \left( \frac{\partial z}{\partial x} \right)^2 + \left( \frac{\partial z}{\partial y} \right)^2} \, dx \, dy \]Here, \( \frac{\partial z}{\partial x} = 2x \) and \( \frac{\partial z}{\partial y} = 2y \) reflect how the surface changes with each small shift in \(x\) or \(y\), and the integrals measure these changes over the defined region.