Question

Use a change of variables to evaluate the following integrals. \(\iiint_{D} d V ; D\) is bounded by the upper half of the ellipsoid \(x^{2} / 9+y^{2} / 4+z^{2}=1\) and the \(x y\) -plane. Use \(x=3 u\) \(y=2 v, z=w\)

Short answer

In summary: Given the equation of the upper half of the ellipsoid and the xy-plane as the other boundary, the change of variables transformation was used to evaluate the volume of the region D. The Jacobian of the transformation was calculated to be 6, and the new bounds of integration were determined using cylindrical coordinates. Finally, the triple integral was evaluated, yielding a final result of \(\frac{1}{2}\) for the volume of the region D.

Step-by-step solution

Find the Jacobian of the change of variables transformation

First, let's find the Jacobian of the transformation, that is, the determinant of the Jacobian matrix. Given the transformations \(x = 3u, y=2v, z=w\). The Jacobian matrix for this set of transformations is: $$ \begin{bmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} & \frac{\partial x}{\partial w} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} & \frac{\partial y}{\partial w} \\ \frac{\partial z}{\partial u} & \frac{\partial z}{\partial v} & \frac{\partial z}{\partial w} \end{bmatrix} = \begin{bmatrix} 3 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 1 \end{bmatrix} $$ Now calculate the determinant of this matrix to find the Jacobian. $$ J(u, v, w)=\det\begin{bmatrix} 3 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 1 \end{bmatrix} = (3)(2)(1) = 6 $$

Transform the ellipsoid equation and find bounds of integration

Now, let's transform the ellipsoid equation using the given change of variables. $$ \frac{(3u)^2}{9} + \frac{(2v)^2}{4} + w^2 = 1 $$ Simplify the equation: $$ u^2 + v^2 + w^2 = 1 $$ Since we are looking at the upper half of the ellipsoid, we will only consider \(w \geq 0\). The bounds of integration can be determined using cylindrical coordinates as \(u^2+v^2\leq1\), \(w\geq0\). Hence, the limits for u, v, and w are: $$ -1\leq u\leq 1\quad -\sqrt{1 - u^2}\leq v\leq \sqrt{1 - u^2}\quad 0\leq w\leq \sqrt{1 - u^2 - v^2} $$

Compute the triple integral using the transformation

Now that we have the Jacobian and the new bounds of integration, we can compute the integral. $$ \iiint_D dV = \iiint_D J(u,v,w) \,du\,dv\,dw = 6 \cdot \int_{-1}^{1}\int_{-\sqrt{1-u^2}}^{\sqrt{1-u^2}}\int_{0}^{\sqrt{1-u^2-v^2}}\,dw\,dv\,du $$ First, evaluate the integral with respect to \(w\). $$ 6 \cdot \int_{-1}^{1}\int_{-\sqrt{1-u^2}}^{\sqrt{1-u^2}}\left[\frac{1}{2}w^2\right]_{0}^{\sqrt{1-u^2-v^2}}\,dv\,du = 3\cdot \int_{-1}^{1}\int_{-\sqrt{1-u^2}}^{\sqrt{1-u^2}}\left(1-u^2-v^2\right)\,dv\,du $$ Now evaluate the integral with respect to \(v\). $$ 3 \cdot \int_{-1}^{1}\left[\frac{1}{3}v(1-u^2-v^2)\right]_{-\sqrt{1-u^2}}^{\sqrt{1-u^2}}\,du = 3 \cdot \int_{-1}^{1}\frac{2}{3}\left(1-u^2\right)^{3/2}\,du $$ Finally, evaluate the integral with respect to \(u\). $$ 2 \cdot \left[\frac{1}{4}\left(1-u^2\right)^{2}\right]_{-1}^{1} = 2 \cdot \frac{1}{4}(1 - (-1))^2 = 2 \cdot \frac{1}{4}\cdot 1 = \frac{1}{2} $$ The value of the integral \(\iiint_D dV\) is \(\frac{1}{2}\), which is the volume of the region D.

Key concepts

Jacobian Matrix

When tackling multi-variable integrals that involve a change of variables, the Jacobian matrix is crucial. Simply put, it helps us understand how the variables change in relation to each other. For example, we see this when converting from one coordinate system to another.
The Jacobian matrix consists of partial derivatives, giving us a map of how each new variable influences the old ones.In our problem, we're making a strategic switch from the space defined by \(x\), \(y\), and \(z\) to a new space using variables \(u\), \(v\), and \(w\). Our transformation equations are:

• \(x = 3u\)
• \(y = 2v\)
• \(z = w\)

This gives us the Jacobian matrix:\[\begin{bmatrix}\frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} & \frac{\partial x}{\partial w} \\frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} & \frac{\partial y}{\partial w} \\frac{\partial z}{\partial u} & \frac{\partial z}{\partial v} & \frac{\partial z}{\partial w} \\end{bmatrix} = \begin{bmatrix}3 & 0 & 0 \0 & 2 & 0 \0 & 0 & 1\end{bmatrix}\]Calculating the determinant, we get \(6\). This determinant, called the Jacobian, acts as a scale factor that we use when changing variables in an integral.
It's indispensable, ensuring we're on the right track by adjusting for the scale change during transformation.

Triple Integrals

A triple integral is like taking regular integration a few steps further by finding the volume under a surface in three-dimensional space. When you evaluate a triple integral, you're trying to solve for a volume or mass that has varying density over the space.In our specific scenario, the aim is to compute the volume bounded by the specified ellipsoid. Once we've set up a change of variables and found the Jacobian, we now apply it to our triple integral.The steps are as follows:

• Find the Jacobian scaling factor, already determined as \(6\).
• Transform the triple integral using this Jacobian.

This modification adapts the current integral into a simplified form, considering the ellipsoid's new boundaries in the function of \(u\), \(v\), and \(w\):\[6 \cdot \int_{-1}^{1}\int_{-\sqrt{1-u^2}}^{\sqrt{1-u^2}}\int_{0}^{\sqrt{1-u^2-v^2}} \,dw \,dv \,du\]By working through each layer of this integral, starting with \(w\), then \(v\), and finally \(u\), we methodically evaluate the volume, reaching the straightforward outcome, which reflects the entire bounded region.

Ellipsoid Transformation

The process of transforming an ellipsoid into a more manageable form through a change of variables is a valuable technique that simplifies integration. The ellipsoid in this problem:\[\frac{x^2}{9} + \frac{y^2}{4} + z^2 = 1\]represents a three-dimensional shape, elongated along the \(x\), \(y\), and \(z\) axes.By introducing new variables \(u\), \(v\), and \(w\), we are effectively conducting a scaling that normalizes this ellipsoid into a unit sphere:\[u^2 + v^2 + w^2 = 1\]This conversion simplifies the limits of integration, making them easier to work with:

• \(-1 \leq u \leq 1\)
• \(-\sqrt{1-u^2} \leq v \leq \sqrt{1-u^2}\)
• \(0 \leq w \leq \sqrt{1-u^2-v^2}\)

Having redefined the ellipsoid as a sphere allows us to focus on only the region of interest— the upper half above the \(xy\)-plane. This transition is pivotal as it offers an approach to break down any complex, asymmetric shape into a more straightforward geometric form when performing integrations.