Question

Find the area of the indicated surface. Make a sketch in each case. The part of \(z=9-x^{2}-y^{2}\) above the plane \(z=5\).

Short answer

The area of the surface above the plane is approximately 68.56 square units.

Step-by-step solution

Determine the Region Above the Plane

First, we need to find the region of the surface \(z = 9 - x^2 - y^2\) that lies above the plane \(z = 5\). Set \(9 - x^2 - y^2 = 5\) and solve for \(x^2 + y^2\) to find the boundary of this region.\\[\begin{align*} 9 - x^2 - y^2 &= 5 \ x^2 + y^2 &= 4 \ \end{align*}\]This equation represents a circle with radius 2 in the \(xy\)-plane.

Set Up the Integral for Surface Area

The surface area of the section of \(z = 9 - x^2 - y^2\) above the plane can be found using the surface area integral:\[ S = \iint_R \sqrt{1 + \left(\frac{\partial z}{\partial x}\right)^2 + \left(\frac{\partial z}{\partial y}\right)^2} \ dA \]Calculate \(\frac{\partial z}{\partial x}\) and \(\frac{\partial z}{\partial y}\). \[\begin{align*}\frac{\partial z}{\partial x} &= -2x, \\frac{\partial z}{\partial y} &= -2y\end{align*}\]

Evaluate the Surface Area Integral

Substitute the derivatives into the integrand:\[ \sqrt{1 + (-2x)^2 + (-2y)^2} = \sqrt{1 + 4x^2 + 4y^2} = \sqrt{1 + 4(x^2 + y^2)} \]With \( x^2 + y^2 = r^2 \), the expression becomes \( \sqrt{1 + 4r^2} \).Convert to polar coordinates where \(x = r\cos(\theta)\) and \(y = r\sin(\theta)\), and \( dA = r\,dr\,d\theta \).\[ S = \int_0^{2\pi} \int_0^2 \sqrt{1 + 4r^2} \ r \, dr \, d\theta \]

Perform the Integration

Complete the integration over \(r\) and \(\theta\):First, integrate with respect to \(r\):\[ \int_0^2 \sqrt{1 + 4r^2} \ r \, dr \]Use the substitution \(u = 1 + 4r^2\), \(du = 8r \, dr\), translating the limits from \(r=0\) to \(r=2\) results in \(u=1\) to \(u=17\). This gives:\[ \frac{1}{8} \int_1^{17} u^{1/2} \, du = \frac{1}{8} \cdot \frac{2}{3} \left[u^{3/2}\right]_1^{17} = \frac{1}{12} \, \left[17^{3/2} - 1^{3/2}\right] \]Evaluate this integral and then multiply by the integral over \(\theta\):\[ S = \left(\frac{1}{12} \cdot (17^{3/2} - 1)\right) \cdot 2\pi \]

Compute the Final Area

Calculate \(17^{3/2}\) to simplify the expression:\[ 17^{3/2} = \sqrt{17^3} = \sqrt{4913} \approx 131.111 \S \approx \frac{1}{12} \cdot (131.111 - 1) \cdot 2\pi \S \approx \frac{1}{12} \cdot 130.111 \cdot 6.283 \]Finally, calculate the numerical value:\[ S \approx 68.56 \]

Sketch the Surface and Region

Draw a sketch of the paraboloid \(z = 9 - x^2 - y^2\) with the plane \(z=5\). The intersection forms a circle of radius 2 in the \(xy\)-plane, representing the region over which the surface area was calculated.

Key concepts

Surface Integral

To determine the surface area of a portion of a surface in space, we commonly employ the concept of a surface integral. This can be thought of as an extension of a double integral that allows us to calculate the total area of a curved surface. In this exercise, our surface is defined by a paraboloid: \(z = 9 - x^2 - y^2\). We want to find the area of this surface above the plane \(z = 5\).
To set up a surface integral for this problem, we need to calculate partial derivatives and use them in the formula:

• First, substitute the derivatives \(\frac{\partial z}{\partial x} = -2x\) and \(\frac{\partial z}{\partial y} = -2y\) into \(\sqrt{1 + \left(\frac{\partial z}{\partial x}\right)^2 + \left(\frac{\partial z}{\partial y}\right)^2}\).
• This results in \(\sqrt{1 + 4x^2 + 4y^2}\) which simplifies to \(\sqrt{1 + 4(x^2 + y^2)}\).

Integrating this expression over the appropriate region will yield the surface area. This integral tells us the sum of all "tiny" surface patches over the paraboloid section.

Polar Coordinates

Polar coordinates are a transformation of the coordinate system which uses the distance from the origin \(r\) and the angle \(\theta\) instead of \(x\) and \(y\). This is especially useful for regions with circular symmetry, such as the region bounded by the circle formed at the intersection of the paraboloid and plane, \(x^2 + y^2 = 4\).
In polar coordinates, we substitute:

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

By using polar coordinates in this surface integral, the limits of integration become simpler. \(r\) ranges from 0 to 2, the radius of our circle, and \(\theta\) from 0 to \(2\pi\), completing a full rotation around the circle.

Paraboloid

A paraboloid is a three-dimensional surface resembling a parabolic shape. In this exercise, we are dealing with an elliptic paraboloid of the form \(z = 9 - x^2 - y^2\).
The term "elliptic" means that any cross-section parallel to the \(xy\)-plane will yield an ellipse or, as in our problem, potentially a circle.
This shape always forms a bowl-like structure opening upwards or downwards. The section of the surface we are interested in is the part above the plane \(z = 5\).
By setting \(9 - x^2 - y^2 = 5\), we determined the boundary: \(x^2 + y^2 = 4\), representing a circle. Therefore, this emphasizes the importance of understanding the geometric shape, as it directly informs the setup and evaluation of integrals.

Integration Techniques

Integration techniques involved in this problem include substitution and breaking down integrals into simpler components. Here, our surface integral depends on integrating \(\sqrt{1 + 4r^2} \cdot r\).
To simplify the process:

• Use substitution: Let \(u = 1 + 4r^2\), which simplifies the expression within the square root. Calculate \(du = 8r\,dr\).
• Adjust limits accordingly, converting \(r=0\) to \(r=2\) into \(u=1\) and \(u=17\).

These techniques also involve calculating integrals over specified bounds for \(r\) and \(\theta\), evaluating the expression with respect to \(u\) by determining anti-derivatives, and then multiplying results by constants that arise from the change in variables. This demonstrates the methods of simplifying complex integrals in surface area calculations.