Question

Use a change of variables to evaluate the following integrals. \(\iiint_{D} z d V ; D\) is bounded by the paraboloid \(z=16-x^{2}-4 y^{2}\) and the \(x y\) -plane. Use \(x=4 u \cos v, y=2 u \sin v, z=w\)

Short answer

To summarize, we were asked to evaluate the triple integral of a function over a region bounded by a paraboloid and the xy-plane. We performed a change of variables using the given transformation and found the Jacobian determinant. After finding the limits of integration and setting up the integral, we evaluated it step by step. The final value of the triple integral is \(\frac{128\pi}{5}\).

Step-by-step solution

Change of Variables

The first step is to rewrite the given transformation in the form of Jacobian. Transformation: \(x=4u\cos{v}\), \(y=2u\sin{v}\), \(z=w\) To find the Jacobian, we write the partial derivatives in the following matrix: \(J(u,v,w) = \begin{bmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} & \frac{\partial x}{\partial w} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} & \frac{\partial y}{\partial w} \\ \frac{\partial z}{\partial u} & \frac{\partial z}{\partial v} & \frac{\partial z}{\partial w} \end{bmatrix} = \begin{bmatrix} 4\cos{v} & -4u\sin{v} & 0 \\ 2\sin{v} & 2u\cos{v} & 0 \\ 0 & 0 & 1 \end{bmatrix}\) Now, we find the determinant of the matrix: \(|J(u,v,w)| = \begin{vmatrix} 4\cos{v} & -4u\sin{v} & 0 \\ 2\sin{v} & 2u\cos{v} & 0 \\ 0 & 0 & 1 \end{vmatrix} = 8u\cos^2{v} + 8u\sin^2{v} = 8u\) We also need to rewrite the equation of the paraboloid and find the limits of integration. \(w = 16 - x^2 - 4y^2 = 16 - (4u\cos{v})^2 - 4(2u\sin{v})^2 = 16 - 16u^2\cos^2{v} - 16u^2\sin^2{v} = 16(1-u^2)\)

Setting up the limits of integration

The region D is bounded by the paraboloid and the \(xy\)-plane. The \(xy\)-plane can be represented as \(z=0\). So this is the lower limit for the z-axis (w-axis in \(uvw\) coordinates). The upper limit will be the paraboloid. Now we have the following limits of integration for our coordinates: 1. \(0 \le u \le 1\) 2. \(0 \le v \le 2\pi\) 3. \(0 \le w \le 16(1-u^2)\)

Setting up the integral

Now we have all the necessary information to set up our integral. \(\iiint_{D} zdV = \int_0^1 \int_0^{2\pi} \int_0^{16(1-u^2)} w\cdot |J(u,v,w)| dw dv du = \int_0^1\int_0^{2\pi}\int_0^{16(1-u^2)} w\cdot 8u dw dv du\)

Evaluating the integral

Evaluate the integral step by step: First, integrate with respect to w: \(\int_0^{16(1-u^2)} 8uw dw = \left[4uw^2\right]_0^{16(1-u^2)} = 64u(1-u^2)^2\) Integrate with respect to v: \(\int_0^{2\pi} 64u(1-u^2)^2 dv = 128\pi u(1-u^2)^2\) Finally, integrate with respect to u: \(\int_0^1 128\pi u(1-u^2)^2 du = \frac{128\pi}{5}\) Thus, the value of the given triple integral is: \(\iiint_{D} zdV = \frac{128\pi}{5}\).

Key concepts

Jacobian Determinant

The Jacobian Determinant is a crucial tool when it comes to changing variables in multiple integrals. If you have a transformation of coordinates like \( x = 4u\cos{v} \), \( y = 2u\sin{v} \), and \( z = w \), you can use the Jacobian to adjust the integration measure for these new variables.

This determinant helps quantify how much the area or volume is 'stretched' by the transformation. Calculating the Jacobian involves forming a matrix of partial derivatives that describe how each new variable affects the original variable. For the transformation here, the matrix is
\[ J(u,v,w) = \begin{bmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} & \frac{\partial x}{\partial w} \ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} & \frac{\partial y}{\partial w} \ \frac{\partial z}{\partial u} & \frac{\partial z}{\partial v} & \frac{\partial z}{\partial w} \end{bmatrix} = \begin{bmatrix} 4\cos{v} & -4u\sin{v} & 0 \ 2\sin{v} & 2u\cos{v} & 0 \ 0 & 0 & 1 \end{bmatrix} \]

The determinant of this matrix, \( |J(u,v,w)| \), equals \( 8u \), which represents how the volume element \( dV \) transforms in \( uvw \) space.

Triple Integral

A triple integral allows you to integrate functions over three-dimensional regions. When you're dealing with a region like the one bounded by a paraboloid, the triple integral can give you quantities like volume or mass if the function represents density.

In an exercise involving triple integrals and change of variables, such as this, you rewrite the integral in your new coordinate system. The region \( D \) transformed into \( uvw \) coordinates, helps us to set up the triple integral as
\[ \iiint_{D} z dV = \int_0^1 \int_0^{2\pi} \int_0^{16(1-u^2)} w\cdot |J(u,v,w)| dw dv du \]

Here you integrate sequentially over \( w \), \( v \), and \( u \). Each step simplifies the integral by focusing on one variable at a time, using the limits set by the transformed paraboloid.

Integration Limits

Setting the correct limits for integration defines the boundaries of the region over which you're integrating. They're critical when dealing with a transformed region like a paraboloid.

For our transformation, the region bounded by the paraboloid is converted into \( uvw \) space. The limits for each integral are based on the geometry of the region:

• \( 0 \leq u \leq 1 \): \( u \) represents a radial boundary.
• \( 0 \leq v \leq 2\pi \): \( v \) represents rotational symmetry, covering a full circle.
• \( 0 \leq w \leq 16(1-u^2) \): \( w \) describes heights determined by the paraboloid equation.

These limits ensure that you only integrate over the specified "slice" of space under the paraboloid.

Paraboloid Equation

The equation of a paraboloid defines a symmetric, bowl-shaped surface. In this problem, the paraboloid is given by the equation \( z = 16 - x^2 - 4y^2 \). This equation describes the shape and extends equidistantly in all directions from the peak at \( z = 16 \) in the positive z-direction until it meets the xy-plane.

Once transformed to \( uvw \) coordinates, it maintains the shape but lets us utilize symmetries and simpler bounds. The transformed equation \( w = 16(1-u^2) \) describes the upper boundary (the lid) in new coordinate space. The bottom boundary is the xy-plane, represented by \( z = 0 \) or \( w = 0 \).

Understanding the nature of this shape helps set up proper boundaries and streamline the calculation by exploiting symmetry, significantly simplifying integration over such regions.