Problem 256 - Advanced Calculus

Question

Compute the area of the paraboloid given by the equation \(z=x^{2}+y^{2}\), with \(0 \leq z \leq 2\)

Step-by-Step Solution

Answer

The area of the paraboloid given by the equation \(z = x^2 + y^2\) with \(0 \leq z \leq 2\) is \(A = \dfrac{\pi}{6} (5\sqrt{5} - 1)\).

Step 1: Find the partial derivatives of z

The first step is to find the partial derivatives of \(z = x^2 + y^2\) with respect to \(x\) and \(y\). \(\dfrac{\partial z}{\partial x} = 2x\) \(\dfrac{\partial z}{\partial y} = 2y\)

Step 2: Determine the projection D of the surface onto the xy-plane

We have \(0 \leq z \leq 2\), so the equation becomes: \(\sqrt{x^2 + y^2} \leq z \leq 2\) For \(z = 2\), we have: \(x^2 + y^2 = 2\) which is a circle of radius \(\sqrt{2}\) in the xy-plane, and for \(z = 0\): \(x^2 + y^2 = 0\) So, our projection is a circular region with radius \(\sqrt{2}\) in the xy-plane.

Step 3: Convert the integral into polar coordinates

Since we have a circular region, it's more convenient to work in polar coordinates. We'll convert the partial derivatives and dA to polar coordinates next. \(\dfrac{\partial z}{\partial x} = 2r\cos\theta\) \(\dfrac{\partial z}{\partial y} = 2r\sin\theta\) and \(dA = r\, dr\, d\theta\).

Step 4: Set up and solve the integral

Now we can set up the integral for finding the area of the paraboloid using the polar coordinates. \[A = \iint_D \sqrt{1 + \left(2r\cos\theta\right)^2 + \left(2r\sin\theta\right)^2} \, r\, dr\, d\theta\] We have limits on r and θ since we are in polar coordinates: \(0 \leq r \leq \sqrt{2}\) and \(0 \leq \theta \leq 2\pi\). \[A = \int_0^{2\pi} \int_0^{\sqrt{2}} \sqrt{1 + (2r\cos\theta)^2 + (2r\sin\theta)^2} \, r\, dr\, d\theta\] \[A = \int_0^{2\pi} \int_0^{\sqrt{2}} \sqrt{1 + 4r^2} \, r\, dr\, d\theta\] Now, let's solve the inner integral first. \[\int_0^{\sqrt{2}} \sqrt{1 + 4r^2} \, r\, dr\] This integral is solvable using substitution. Let \(u = 1 + 4r^2\), therefore \(\dfrac{du}{dr} = 8r\) and so \(dr = \dfrac{1}{8r}\, du\). The integral becomes: \[\int \sqrt{u} \, r\, \dfrac{1}{8r}\, du = \dfrac{1}{8} \int_0^5 u^{1/2}\, du\] Now, we can evaluate this integral: \[\dfrac{1}{8} \left[\dfrac{2}{3}u^{3/2}\right]_0^5 = \dfrac{1}{12} (5\sqrt{5} - 1)\] Now, we can evaluate the outer integral with respect to θ: \[A = \int_0^{2\pi} \left(\dfrac{1}{12} (5\sqrt{5} - 1)\right) d\theta\] \[A = \left(\dfrac{1}{12} (5\sqrt{5} - 1)\right) \int_0^{2\pi} d\theta = \left(\dfrac{1}{12} (5\sqrt{5} - 1)\right) (2\pi)\] Hence, the area of the paraboloid is: \[A = \dfrac{\pi}{6} (5\sqrt{5} - 1)\]

Key Concepts

Partial Derivatives

Partial derivatives are a fundamental concept in calculus, particularly useful when dealing with functions of multiple variables. In this exercise, we start by considering the paraboloid described by the equation \( z = x^2 + y^2 \). The goal is to find how this surface changes in relation to each individual variable, \( x \) and \( y \), while keeping the other constant.

To achieve this, we compute the partial derivatives:

• First, with respect to \( x \), you treat \( y \) as a constant resulting in: \( \dfrac{\partial z}{\partial x} = 2x \)
• Similarly, with respect to \( y \), treating \( x \) as constant: \( \dfrac{\partial z}{\partial y} = 2y \)

These derivatives show the rate of change of the height of the paraboloid as we move along \( x \)-axis and \( y \)-axis respectively. Understanding these rates of change is crucial when we later calculate the surface area.

Polar Coordinates

Polar coordinates offer a convenient way to handle problems involving circular regions, like our projected area on the xy-plane. Here, we convert the problem into polar coordinates as the surface has circular symmetry.

In polar coordinates, any point \((x, y)\) can be described by \((r, \theta)\), where:

• \(r\) is the radial distance from the origin, and
• \(\theta\) is the angle from the positive x-axis.

The equations for \(x\) and \(y\) in terms of \(r\) and \(\theta\) are therefore \(x = r\cos\theta\) and \(y = r\sin\theta\).

Derivatives are also transformed:

• \( \dfrac{\partial z}{\partial x} = 2r\cos\theta \)
• \( \dfrac{\partial z}{\partial y} = 2r\sin\theta \)

Additionally, the differential area element, initially \(dA\), becomes \(r\, dr\, d\theta\) to accommodate the integration over the circular area. This conversion simplifies the integral setup required for surface area calculations.

Integral Calculus

Integral calculus enables us to find the total surface area of a complex 3-dimensional shape like a paraboloid. Here, by integrating the square root of the sum of these squared partial derivatives plus one, we calculate the surface area.

The key to setting up the integral is the formula for surface area in terms of partial derivatives, leading us to: \[A = \iint_D \sqrt{1 + \left(\left(\frac{\partial z}{\partial x}\right)^2 + \left(\frac{\partial z}{\partial y}\right)^2 \right)} \, dA\]

This integrand represents the local infinitesimal surface area, with \(dA\) changing according to the coordinate system used—in our case becoming \(r\, dr\, d\theta\) using polar coordinates.
To evaluate, we integrate first with respect to \(r\), then \(\theta\). This step-by-step approach is typical of integral calculus and underscores its power in summing infinitesimally small areas to derive exact total areas.

Double Integral

Double integrals are a core part of multivariable calculus, allowing us to compute quantities like area and volume over regions in two-dimensional space. In this example, we use a double integral to find the paraboloid's surface area.

The double integral is:\[A = \int_0^{2\pi} \int_0^{\sqrt{2}} \sqrt{1 + 4r^2} \, r\, dr\, d\theta\]

This integral is evaluated over a region defined in polar coordinates:

• \(r\) varies from 0 to \(\sqrt{2}\), covering the radial dimension of the circle.
• \(\theta\) ranges from 0 to \(2\pi\), covering the entire angular span.

The double integral effectively "adds up" all the local areas of tiny patches on the surface to yield the total surface area. Evaluating this integral with careful application of techniques like substitution and applying these limits gives us the final result for the area. This approach showcases the power of double integrals in handling geometrically complex cases.