Question

As needed, use a computer to plot graphs and to check values of integrals. (a) Find the area of the surface \(z=1+x^{2}+y^{2}\) inside the cylinder \(x^{2}+y^{2}=1\). (b) Find the volume inside the cylinder between the surface and the \((x, y)\) plane. Use cylindrical coordinates.

Short answer

(a) Surface area is \frac{\pi}{6} (5\sqrt{5} - 1)\. (b) Volume is \2\pi/3\.

Step-by-step solution

Describe the given surface and cylinder

The surface is given by the equation \(z = 1 + x^2 + y^2\) and the cylinder is described by the equation \(x^2 + y^2 = 1\). We need to find the area and volume inside this surface bounded by the cylinder.

Set up the integral for the surface area

The surface area element in rectangular coordinates is given by \(dS = \sqrt{1 + \left(\frac{\partial z}{\partial x}\right)^2 + \left(\frac{\partial z}{\partial y}\right)^2} \, dx \, dy\). Here, \(\frac{\partial z}{\partial x} = 2x\) and \(\frac{\partial z}{\partial y} = 2y\).

Simplify and convert to cylindrical coordinates

In cylindrical coordinates, \(x = r\cos\theta\), \(y = r\sin\theta\), and the surface equation becomes \(z = 1 + r^2\). The surface area element becomes \(dS = \sqrt{1 + 4r^2} \, r \, dr \, d\theta\).

Evaluate the surface area integral

Evaluate the integral \(\int_{0}^{2\pi} \int_{0}^{1} \sqrt{1 + 4r^2} \, r \, dr \, d\theta \). Solve \(\int_{0}^{1} \, r\sqrt{1 + 4r^2} \, dr\) first, then integrate the result with respect to \(\theta\) from \(0\) to \(2\pi\).

Find the volume using cylindrical coordinates

The volume is given by the integral \(\int_{0}^{2\pi} \int_{0}^{1} \int_{0}^{1+r^2} r \, dz \, dr \, d\theta\). Integrate with respect to \(z\), then \(r\), and finally \(\theta\) from \(0\) to \(2\pi\).

Compute each integral

For the surface area, we have \(\int_{0}^{2\pi} d\theta \, \int_{0}^{1} r \sqrt{1 + 4r^2} \, dr\). For the volume, \(\int_{0}^{2\pi} \int_{0}^{1} r [1 + r^2] \, dr \, d\theta\). Solve these integrals separately.

Key concepts

surface area calculation

To calculate the surface area of a given surface within a specified region, we can use surface integrals. For a function described by \(z = f(x, y)\), the surface area element is computed using: \[dS = \sqrt{1 + \left(\frac{\partial z}{\partial x}\right)^2 + \left(\frac{\partial z}{\partial y}\right)^2} \, dx \, dy\].
Given \(z = 1 + x^2 + y^2\), we calculate the partial derivatives: \(\frac{\partial z}{\partial x} = 2x\) and \(\frac{\partial z}{\partial y} = 2y\).
Substituting these into the surface area element formula, we get:
\[\sqrt{1 + 4x^2 + 4y^2}\] dx dy.
This computation will give the surface area across the given region when integrated over the cylinder \(x^2 + y^2 = 1\). By converting to cylindrical coordinates for simplicity, we set \(x = r\cos\theta\), \(y = r\sin\theta \), and the surface function becomes \(z = 1 + r^2\). Thus, the surface integral in cylindrical coordinates is \[\int_{0}^{2\pi} \int_{0}^{1} \sqrt{1 + 4r^2} \, r \, dr \, d\theta\].
Remember, integrating \[\int_{0}^{1} \ r \sqrt{1 + 4r^2} \, dr\], and then over \[\int_{0}^{2\pi} d\theta\] gives the total surface area calculation.

volume integrals

Volume integrals help compute the volume of a given region in three-dimensional space. The volume bounded by the surface \(z = f(x, y)\) and a region can be found by evaluating:
\[\int_{\Omega} (f(x, y) - g(x, y)) dA\],
where \(dA\) represents the area element in the xy-plane.
For the function \(z = 1 + x^2 + y^2\) inside the cylinder \(x^2 + y^2 = 1\), we can convert to cylindrical coordinates for simplicity.
In cylindrical coordinates, \[\int_{0}^{2\pi} \int_{0}^{1} \int_{0}^{1+r^2} r \, dz \, dr \, d\theta\] will represent the volume bounded by the surface and the xy-plane within the cylinder.
Here:

• Integrate first with respect to \(z\), yielding \[ \int_{0}^{1+r^2} dz = 1+ r^2\].
• Next, integrate with respect to \(r\).
• Finally, integrate with respect to \(\theta\) over \(0\) to \(2\pi\).

The integration order and limits are crucial in determining the accurate volume. Incorporating cylindrical coordinates simplifies these integrals significantly.

conversion to cylindrical coordinates

Converting to cylindrical coordinates can simplify integrals over regions and volumes involving circular symmetry. In cylindrical coordinates, any point in space is given by:

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

The Jacobian determinant for this transformation introduces an additional \(r \, dr \, d\theta\) term in integrals.
Another advantage is the transformation of the surfaces or volumes with circular bounds. For example, the cylinder \(x^2 + y^2 = 1\) becomes \(r = 1\) in cylindrical coordinates simplifying the boundaries.
Given a surface like \(z = 1 + x^2 + y^2\), converting to cylindrical coordinates using \(r\) and \(\theta\) transforms it into \[ z = 1 + r^2. \] This transforms the surface area integrand to \[ \sqrt{1 + 4r^2} \, r \, dr \, d\theta\], thus simplifying the integral calculations.
These transformations often make otherwise complex integrals not only solvable but also far simpler.