Question

Find the area of the paraboloid \(x^{2}+y^{2}=z\) inside the cylinder \(x^{2}+y^{2}=9\)

Short answer

The area is \(9\pi\).

Step-by-step solution

Understand the region of integration

Identify the region of integration. We need to find the area of the paraboloid given by the equation \(x^2 + y^2 = z\) that is inside the cylinder \(x^2 + y^2 = 9\). The radius of the cylinder is 3 (since \(9 = 3^2\)). The range for radius \(r\) in polar coordinates is from 0 to 3.

Set up the integral in cylindrical coordinates

Switch to polar coordinates where \(x = r \, \text{cos} \, \theta\) and \(y = r \, \text{sin} \, \theta\). The equation of the paraboloid becomes \(r^2 = z\). Use the area differential in cylindrical coordinates \(dA = r \, dr \, d\theta\).

Define the limits of integration

For the given cylinder, the radius \(r\) ranges from 0 to 3 and the angle \(\theta\) ranges from 0 to \(2\pi\). Therefore, \(r\text{-range: } [0, 3]\) and \(\theta\text{-range: } [0, 2\pi]\).

Integrate with respect to \(r\)

Set up and solve the integral. The height of the surface at radius \(r\) is \(z = r^2\). The total area \(A\) is an integral over the surface: \[ A = \int_{0}^{2\pi} \int_{0}^{3} r \, \text{d}r \, \text{d}\theta \].

Calculate the integral

\( \int_{0}^{2\pi} \text{d}\theta = 2\pi \)\( \int_{0}^{3} r \, \text{d}r = \left[ \frac{r^2}{2} \right]_{0}^{3} = \frac{9}{2} \).Multiplying these results together gives the total area: \[ A = 2\pi \times \frac{9}{2} = 9\pi. \]

Key concepts

Multivariable Calculus

Multivariable calculus deals with functions of multiple variables. These functions depend on more than one input, which means you have to consider changes in each variable's effect. For instance, when calculating the area of a surface in three dimensions, it becomes essential to understand not just the \(x\) and\(y\) coordinates, but how a change in these affects the \(z\) coordinate as well.
This is particularly useful when finding areas or volumes of complex shapes, like a paraboloid. It’s the foundation which makes other techniques, like double integration, possible and comprehensible.
Following the step-by-step solution, we apply this understanding to integrate over a region in space to find the area of the paraboloid.

Cylindrical Coordinates

Switching to cylindrical coordinates can simplify problems involving symmetrical shapes like cylinders or paraboloids. In cylindrical coordinates, the position of a point in 3D space is determined using a radius \(r\), angle \(\theta\), and height \(z\). This system extends polar coordinates into three dimensions:

• \(r\) - radial distance from the origin
• \(\theta\) - angle around the origin
• \(z\) - height

The beauty of cylindrical coordinates lies in how they simplify integration. Instead of working with \(x\) and \(y\), we use \(r\) and \(\theta\). This means the area element \(dA\) in Cartesian coordinates turns into \(r \, dr \, d\theta\) in cylindrical coordinates. The switch makes it easier to set up and evaluate the integral.

Polar Coordinates

Polar coordinates are the backbone of cylindrical coordinates but are used in 2D space. They describe locations using a radius \(r\) and an angle \(\theta\) from a reference direction:

• \(r\) - distance from the origin
• \(\theta\) - angle from the positive \(x\)-axis

Transitioning to polar coordinates can make certain integrals easier to evaluate, especially when the problem exhibits circular or rotational symmetry.
For the paraboloid problem, polar coordinates transform the complex integral into a more manageable form. Hence, \(x\) and \(y\) are converted to \(r \, cos \theta\) and \(r \, sin \theta\). Points within the circle \(x^2 + y^2 = 9\) become \(r^2 = 9\), with \(r\) ranging from 0 to 3 and \(\theta\) from \(0\) to \(2\pi\).

Double Integration

Double integration involves evaluating an integral over a 2-dimensional region. It’s like summing up tiny elements across the surface area. The process is particularly useful for finding areas and volumes:

• \(\int_{a}^{b}\int_{c}^{d} \, f(x, y) \, dx \, dy\) - for functions in Cartesian coordinates
• \(\int_{0}^{2\pi}\int_{0}^{R} \, f(r, \theta) \, r \, dr \, d\theta\) - for functions in polar coordinates

For our paraboloid within a cylinder, double integration translates to using cylindrical or polar coordinates to simplify the bounds and integrand. Here, you integrate \(r \, dr \, d\theta\) over the region defined by \(0 \leq r \leq 3\) and \(0 \leq \theta \leq 2\pi\).
This technique allows summing small areas across the region of interest, providing the total area under the curve efficiently.