Problem 17 - APEX Calculus

Question

Find the area of the given surface over the region \(R\). \(f(x, y)=10-2 \sqrt{x^{2}+y^{2}}\) over \(R,\) bounded by the circle \(x^{2}+y^{2}=25\). (This is the cone with height 10 and base radius 5 ; be sure to compare your result with the known formula.)

Step-by-Step Solution

Answer

The surface area is \(25\pi\sqrt{5}\), which matches the known formula for the cone.

Step 1: Understand the Problem

We need to find the area of the surface generated by the function \(f(x, y) = 10 - 2\sqrt{x^2 + y^2}\) over the region \(R\), which is the area inside the circle \(x^2 + y^2 = 25\). This represents a portion of a cone with height 10 and base radius 5.

Step 2: Identify Relevant Formula for Area of Surface

The formula for the surface area \(A\) of a function \(z = f(x, y)\) over a region \(R\) is given by: \[A = \iint_R \sqrt{1 + \left(\frac{\partial f}{\partial x}\right)^2 + \left(\frac{\partial f}{\partial y}\right)^2} \ dA\]

Step 3: Calculate the Partial Derivatives

Calculate the partial derivatives of \(f\). \(\frac{\partial f}{\partial x} = -\frac{2x}{\sqrt{x^2 + y^2}}\), \(\frac{\partial f}{\partial y} = -\frac{2y}{\sqrt{x^2 + y^2}}\).

Step 4: Substitute Partial Derivatives into Surface Area Formula

Plug these into the surface area formula: \[A = \iint_R \sqrt{1 + \left(-\frac{2x}{\sqrt{x^2 + y^2}}\right)^2 + \left(-\frac{2y}{\sqrt{x^2 + y^2}}\right)^2} \ dA\].Simplify the expression inside the square root: \[\sqrt{1 + \frac{4x^2}{x^2 + y^2} + \frac{4y^2}{x^2 + y^2}} = \sqrt{1 + 4} = \sqrt{5}\].

Step 5: Set Up the Integral in Polar Coordinates

Convert the integral into polar coordinates due to circular symmetry. In polar coordinates, \(x = r\cos\theta\), \(y = r\sin\theta\), and \(dA = r\,dr\,d\theta\).Thus, the region \(R\) is described by \(0 \leq r \leq 5\) and \(0 \leq \theta \leq 2\pi\).The integral becomes: \[A = \int_0^{2\pi} \int_0^5 \sqrt{5} \cdot r \, dr \, d\theta\].

Step 6: Evaluate the Integral

First, compute the inner integral: \[\int_0^5 \sqrt{5} \cdot r \, dr = \sqrt{5} \left[ \frac{r^2}{2} \right]_0^5 = \sqrt{5} \cdot \frac{25}{2} = \frac{25\sqrt{5}}{2}\].Next, compute the outer integral: \[\int_0^{2\pi} \frac{25\sqrt{5}}{2} \, d\theta = \frac{25\sqrt{5}}{2} \cdot [\theta]_0^{2\pi} = \frac{25\sqrt{5}}{2} imes 2\pi = 25\pi\sqrt{5}\].

Step 7: Compare with Known Formula

The known formula for the lateral surface area of a cone is \(\pi r l\), where \(l\) is the slant height. Here, \(r = 5\) and \(l = \sqrt{10^2 + 5^2} = \sqrt{125} = 5\sqrt{5}\), so the area is also \(25\pi\sqrt{5}\), confirming our result.

Key Concepts

Partial Derivatives

Imagine you have a function that changes with respect to multiple variables, say, it depends on both \(x\) and \(y\). Partial derivatives help us understand how this function changes when one of these variables changes, while keeping the other constant. For the function \(f(x, y) = 10 - 2\sqrt{x^2 + y^2}\), partial derivatives were calculated to ensure we understand how this surface tilts or slopes in different directions.

Here is how it works:

• Find \(\frac{\partial f}{\partial x}\). This derivative shows how \(f\) changes with \(x\) when \(y\) is held constant. Using basic calculus rules, we get \(-\frac{2x}{\sqrt{x^2 + y^2}}\).
• Similarly, \(\frac{\partial f}{\partial y}\) shows how \(f\) changes with \(y\), resulting in \(-\frac{2y}{\sqrt{x^2 + y^2}}\).

These partial derivatives become a key part of computing the surface area because they help describe the surface's shape as you move over the region \(R\).

Polar Coordinates

When dealing with areas involving circles or symmetric shapes, like our cone surface, polar coordinates simplify the problem. While Cartesian coordinates use \(x\) and \(y\), polar coordinates use \(r\) and \(\theta\), where \(r\) is the distance from the origin and \(\theta\) is the angle from the positive \(x\)-axis. Here's how we make the conversion:

• \(x = r\cos\theta\)
• \(y = r\sin\theta\)
• The area element \(dA\) then becomes \(r \, dr \, d\theta\)

This conversion is crucial because it fits nicely with the circular nature of the region \(R\), bounded by \(x^2 + y^2 = 25\), or \(r = 5\). By rewriting the surface integral in terms of polar coordinates, we align the math perfectly with the geometry of the problem.

Double Integral

A double integral helps compute areas and volumes under a surface over a given region. In this exercise, using the formula:\[A = \iint_R \sqrt{1 + \left(\frac{\partial f}{\partial x}\right)^2 + \left(\frac{\partial f}{\partial y}\right)^2} \, dA\]We aim to calculate the surface area of the cone. The double integral sums up tiny pieces of area across the entire region \(R\). Here's the simplified process:

• With partial derivatives substituted, the expression inside the integral is condensed to \(\sqrt{5}\).
• Converting the integral to polar coordinates makes the integration more intuitive over a circular region.
• Breaking it into an inner and outer integral: first integrate with respect to \(r\), then with respect to \(\theta\).

By carefully evaluating this double integral, you find the total surface area enclosed by the circular region, ensuring it matches the physical intuition behind a cone's surface.

Conic Sections

Conic sections are curves obtained by slicing a cone. They include circles, ellipses, parabolas, and hyperbolas. Here, we are dealing with a circular base of a cone. Understanding conic sections reinforces the geometry of this exercise, where you consider both the base circle and the lateral surface of the cone itself.

Key insights about the cone:

• The given equation \(x^2 + y^2 = 25\) specifies a circle on the \(xy\)-plane.
• The height of the conic surface is affected by \(z = f(x, y) = 10 - 2\sqrt{x^2 + y^2}\).
• This surface forms a classic cone shape, where the radius at the base is 5 and the peak reaches up to 10.

By understanding how conic sections define shapes, you gain insight into how the surface over region \(R\) behaves, confirming both the calculation of the surface area and the broader geometrical interpretation of the problem.