Problem 45 - Thomas Calculus

Question

Find the area of the portion of the paraboloid \(x=4-y^{2}-z^{2}\) that lies above the ring \(1 \leq y^{2}+z^{2} \leq 4\) in the \(y z\) -plane.

Step-by-Step Solution

Answer

The area is \(\frac{1}{3}\pi (17\sqrt{17} - 5\sqrt{5})\).

Step 1: Understanding the problem

We need to find the surface area of a portion of a paraboloid described by the equation \(x = 4 - y^2 - z^2\). The surface area is limited to where it lies above a ring, defined by the inequality \(1 \leq y^2 + z^2 \leq 4\), in the \(yz\)-plane.

Step 2: Parameterize the surface

To find the surface area, we parameterize the surface using coordinates \(y\) and \(z\). For the paraboloid, let \(\mathbf{r}(y,z) = (4 - y^2 - z^2, y, z)\). This helps in integrating over the region.

Step 3: Calculate the normal vector

The normal vector \(\mathbf{n}\) can be calculated as \(\mathbf{n} = \frac{\partial \mathbf{r}}{\partial y} \times \frac{\partial \mathbf{r}}{\partial z}\). Here, \(\frac{\partial \mathbf{r}}{\partial y} = (-2y, 1, 0)\) and \(\frac{\partial \mathbf{r}}{\partial z} = (-2z, 0, 1)\). Compute \(\mathbf{n} = (1, 2y, 2z)\).

Step 4: Find the magnitude of the normal vector

Find the magnitude of normal vector \(\mathbf{n}\) which is \(\|\mathbf{n}\| = \sqrt{1 + (2y)^2 + (2z)^2} = \sqrt{1 + 4y^2 + 4z^2}\). This magnitude is needed to find the differential surface area element.

Step 5: Set up the integral for the surface area

The surface area \(A\) is given by the integral \(A = \iint_R \|\mathbf{n}\| \, dA\), where \(R\) is the region \(1 \leq y^2 + z^2 \leq 4\). We'll substitute \(\|\mathbf{n}\| = \sqrt{1 + 4(y^2 + z^2)}\) and integrate over the defined region.

Step 6: Convert to polar coordinates

Convert to polar coordinates to simplify the integration. Set \(y = r\cos(\theta)\) and \(z = r\sin(\theta)\). Then, \(dA = r \, dr \, d\theta\) and the region becomes \(1 \leq r^2 \leq 4\), corresponding to \(1 \leq r \leq 2\).

Step 7: Evaluate the integral

Now substitute and integrate: \(A = \int_{0}^{2\pi} \int_{1}^{2} r\sqrt{1 + 4r^2} \, dr \, d\theta\). Solve the inner integral with respect to \(r\) first, followed by the integral with respect to \(\theta\), which involves trigonometric functions and a substitution technique.

Step 8: Perform integration

Start with the substitution \(u = 1 + 4r^2\), then \(du = 8r \, dr\). Substitute back and integrate to simplify the calculating process, solving as \(\int_{1}^{2} r \sqrt{1 + 4r^2} \, dr = \frac{1}{6}(17\sqrt{17} - 5\sqrt{5})\). Multiply by \(2\pi\) after integrating over \(\theta\).

Step 9: Compute final result

Calculate the final result for the surface area: \(A = 2\pi \times \left(\frac{1}{6}(17\sqrt{17} - 5\sqrt{5})\right)\), which results in a numerical computation to find the exact surface area.

Key Concepts

Parameterization of Surfaces

When dealing with surfaces in three-dimensional space, parameterization is a powerful tool. By representing a surface with parameters, we can transform complicated geometric shapes into something easier to work with.
For the given paraboloid, expressed as a function of the coordinates \( y \) and \( z \), we have \( x = 4 - y^2 - z^2 \). This expression maps out how the surface bends and curves in space. To parameterize the surface in terms of \( y \) and \( z \), we define the vector function:

• \( \mathbf{r}(y, z) = (4 - y^2 - z^2, y, z) \)

Through parameterization, we streamline the process of applying calculus to surfaces, making it simpler to calculate things like surface area or analyze properties of the surface. By having the surface in a vector form, any operations needed for integration become more direct, allowing us to focus our efforts on the integration techniques without worrying about the geometry's complexity itself.

Polar Coordinates Integration

Converting surface integration into polar coordinates can simplify the process considerably, especially for regions with circular symmetry. In our problem, the region where the paraboloid lies is described by the ring \(1 \leq y^2 + z^2 \leq 4\). This not only suggests circular symmetry but also encourages the use of polar coordinates.
To convert to polar coordinates, we perform substitutions where:

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

In this setup, \( r \) is the radius, and \( \theta \) is the angle of rotation around the origin. The transformation of the differential area element occurs accordingly, where \( dA = r \, dr \, d\theta \).
For the defined region, constraints in terms of \( r \) become \( 1 \leq r \leq 2 \). This transformation enables us to integrate smoothly over the annular region, with relatively straightforward bounds of integration for both \( r \) and \( \theta \). Integrating in polar coordinates thus offers a more intuitive and manageable path to solving for the surface area.

Normal Vector Calculation

The calculation of the normal vector to a surface is crucial when determining quantities like surface area. Essentially, the normal vector indicates the direction that is perpendicular to the surface at any given point.
For our parameterized surface \( \mathbf{r}(y, z) = (4 - y^2 - z^2, y, z) \), we find the normal vector by computing the cross-product of the partial derivatives:

• \( \frac{\partial \mathbf{r}}{\partial y} = (-2y, 1, 0) \)
• \( \frac{\partial \mathbf{r}}{\partial z} = (-2z, 0, 1) \)

The cross-product, \( \mathbf{n} = \frac{\partial \mathbf{r}}{\partial y} \times \frac{\partial \mathbf{r}}{\partial z} \), gives \( \mathbf{n} = (1, 2y, 2z) \).
The magnitude of the normal vector \( \| \mathbf{n} \| \) is then calculated as \( \sqrt{1 + 4y^2 + 4z^2} \). This magnitude represents the differential element of the surface area when integrated over the proper region. Calculating the normal vector and its magnitude accurately is fundamental in the computation of surface integrations, ensuring the problem is handled with precision.