Question

In Exercises \(15-16,\) special double integrals are presented that are especially well suited for evaluation in polar coordinates. The surface of a right circular cone with height \(h\) and base radius \(a\) can be described by the equation \(f(x, y)=\) \(h-h \sqrt{\frac{x^{2}}{a^{2}}+\frac{y^{2}}{a^{2}}},\) where the tip of the cone lies at \((0,0, h)\) and the circular base lies in the \(x\) -y plane, centered at the origin. Confirm that the volume of a right circular cone with height \(h\) and base radius \(a\) is \(V=\frac{1}{3} \pi a^{2} h\) by evaluating \(\iint_{R} f(x, y) d A\) in polar coordinates.

Short answer

The volume of the cone is \( \frac{1}{3} \pi a^2 h \).

Step-by-step solution

Understanding the Problem

The problem requires us to confirm the volume of a cone using double integrals in polar coordinates. The cone's equation is given and the goal is to evaluate the integral \( \iint_{R} f(x, y) \, d A \) where \( R \) is the region within the circular base.

Convert to Polar Coordinates

In polar coordinates, the transformation is given by \( x = r\cos\theta \) and \( y = r\sin\theta \). The region \( R \) is a circle of radius \( a \) centered at the origin, hence, \( 0 \leq r \leq a \) and \( 0 \leq \theta < 2\pi \). Additionally, the differential area element \( dA \) changes to \( rdrd\theta \).

Express the Function in Polar Coordinates

The function \( f(x, y) \) becomes \( f(r, \theta) = h - h\sqrt{\frac{r^2\cos^2\theta}{a^2} + \frac{r^2\sin^2\theta}{a^2}} = h - h\frac{r}{a} = h\left(1-\frac{r}{a}\right) \).

Set Up the Double Integral in Polar Coordinates

The double integral over the region \( R \) can be written as:\[\int_{0}^{2\pi}\int_{0}^{a} \left( h\left( 1 - \frac{r}{a} \right) \right) r \, dr \, d\theta.\]

Evaluate the Inner Integral

Firstly, we evaluate the inner integral with respect to \( r \):\[\int_{0}^{a} h\left(1 - \frac{r}{a}\right) r \, dr = h \int_{0}^{a} \left(r - \frac{r^2}{a}\right) \, dr.\]Breaking it down further:\[h \left[ \frac{r^2}{2} - \frac{r^3}{3a} \right]_{0}^{a} = h \left( \frac{a^2}{2} - \frac{a^3}{3a} \right) = h \left( \frac{a^2}{2} - \frac{a^2}{3} \right) = h \left( \frac{a^2}{6} \right) \].

Evaluate the Outer Integral

Now, we compute the outer integral:\[\int_{0}^{2\pi} \frac{ha^2}{6} \, d\theta = \frac{ha^2}{6} \int_{0}^{2\pi} \, d\theta = \frac{ha^2}{6} \times 2\pi = \frac{\pi a^2h}{3}.\]

Confirm the Result

The result of \( \frac{\pi a^2h}{3} \) confirms the known formula for the volume of a cone, \( V = \frac{1}{3} \pi a^2 h \).

Key concepts

Understanding Double Integrals

Double integrals are a way of integrating over a two-dimensional area. They are crucial in calculating quantities that spread over a plane, such as the volume under a surface. When we use double integrals, it means we are finding the accumulation of a function over the specified region.
For a function of two variables, like \( f(x, y) \), the double integral is written as \( \iint_{R} f(x, y)\, dA \). This format signifies that we are summarizing the values of the function \( f(x, y) \) across the area \( R \).
Double integrals work by first integrating with respect to one variable while keeping the other constant and then integrating that result with respect to the second variable.
This two-step process allows for the computation of volume if \( f(x, y) \) describes a surface lying above the region \( R \). By using double integrals, we analyze multi-dimensional aspects of functions.

Calculating the Volume of a Cone

The volume of a cone is an important concept in geometry and calculus. A cone is a three-dimensional shape with a circular base and a pointed top called the apex. The formula for the volume of a cone is \( V = \frac{1}{3} \pi a^2 h \), where \( a \) is the radius of the base, and \( h \) is the height.
This relation tells us how much space there is inside the cone. To derive this formula using calculus, we use a double integral. This method involves integrating the function that represents the cone's surface to calculate the total volume.
The geometric interpretation involves imagining stacking infinitesimally thin disks along the height of the cone, each contributing a small volume.
By summing these tiny volumes using integration, we arrive at the familiar formula for the cone's volume.

Coordinate Transformation to Polar Coordinates

Coordinate transformation is converting points from one coordinate system to another. In this problem, we are switching from Cartesian coordinates, \( (x,y) \), to polar coordinates, \( (r,\theta) \).
Polar coordinates describe a point's location based on its angle and distance from the origin. The transformations involve using the formulae \( x = r\cos\theta \) and \( y = r\sin\theta \).
This change is particularly helpful when dealing with circular or symmetric regions, like the base of the cone, because it simplifies the integration process.
The differential area element \( dA \) also changes to \( rdrd\theta \) in polar coordinates. This transformation allows for easier integration when the region \( R \) better fits a circular shape in space.

Integration in Polar Coordinates

Integration in polar coordinates simplifies the process of calculating areas and volumes for circular or symmetric shapes.
In the exercise, after converting to polar coordinates, the task is to integrate the function describing the cone over its basal region. The double integral becomes \( \int_{0}^{2\pi}\int_{0}^{a} \left( h\left( 1 - \frac{r}{a} \right) \right) r \ dr \ d\theta \).
Integrating in polar coordinates involves evaluating the integral with respect to \( r \) first, keeping \( \theta \) constant. After solving the inner integral, we proceed by integrating with respect to \( \theta \) over its applicable range.
This method simplifies the computation by leveraging the region's circular symmetry, thus confirming the cone's volume formula effectively and efficiently.