Question

Use a change of variables to evaluate the following integrals. ${\iiint }_{D}dV;D$ is bounded by the planes $y-2x=0,y-2x=1$ $z-3y=0,z-3y=1,z-4x=0,$ and $z-4x=3$

Short answer

Based on the given planes and the solution provided, the value of the triple integral for the given region D is $\frac{1}{4}$.

Step-by-step solution

Analyze the given planes

We are given six planes: 1. $y-2x=0$ 2. $y-2x=1$ 3. $z-3y=0$ 4. $z-3y=1$ 5. $z-4x=0$ 6. $z-4x=3$ From the planes 1 and 2, let: $u=y-2x$ From the planes 3 and 4, let: $v=z-3y$ From the planes 5 and 6, let $w=z-4x$ Now, we have a system of equations: 1. $u=y-2x$ 2. $v=z-3y$ 3. $w=z-4x$

Find the inverse transformation

From the system of equations, find the inverse transformation by solving for $x,y,z$: 1. $y=u+2x$ 2. $z=v+3y$ 3. $z=w+4x$ Substituting (1) into (2) and (3): $z=v+3(u+2x)$ $z=w+4x$ Now we get: 1. $x=\frac{1}{6}(3v-w)$ 2. $y=u+2x=u+\frac{1}{2}(3v-w)$ 3. $z=\frac{1}{6}(3v+w)$

Find the Jacobian

Compute the Jacobian $J$ of the transformation using the formulas derived in step 2: $J = $$\left|\begin{array}{ccc}\frac{\partial x}{\partial u}& \frac{\partial x}{\partial v}& \frac{\partial x}{\partial w}\\ \frac{\partial y}{\partial u}& \frac{\partial y}{\partial v}& \frac{\partial y}{\partial w}\\ \frac{\partial z}{\partial u}& \frac{\partial z}{\partial v}& \frac{\partial z}{\partial w}\end{array}\right|$$ = $$\left|\begin{array}{ccc}0& \frac{1}{2}& -\frac{1}{6}\\ 1& \frac{1}{2}& -\frac{1}{6}\\ 0& \frac{1}{2}& \frac{1}{6}\end{array}\right|$$$ Using the determinant expansion along the first row, we get: $J=\left|\begin{array}{cc}\frac{1}{2}& -\frac{1}{6}\\ \frac{1}{2}& \frac{1}{6}\end{array}\right|=\frac{1}{36}(2+\frac{1}{3})=\frac{1}{12}$

Set up the triple integral

We now set up the integral with the changed variables: ${\iiint }_{D}dV={\iiint }_{D^{\prime }}|J|dudvdw$ Where $D^{\prime }$ is the new region in the $(u,v,w)$-space. In this space, the bounding planes become: 1. $u=0$ 2. $u=1$ 3. $v=0$ 4. $v=1$ 5. $w=0$ 6. $w=3$

Evaluate the triple integral

Now, evaluate the triple integral: ${\iiint }_{D}dV={\iiint }_{D^{\prime }}|J|dudvdw=\frac{1}{12}{\int }_0^1{\int }_0^1{\int }_0^{3}dudvdw=\frac{1}{12}\cdot (1)\cdot (1)\cdot (3)=\frac{1}{4}$ Thus, the value of the triple integral is $\frac{1}{4}$.

Key concepts

Triple Integral

A triple integral is like an extension of a single or a double integral. It allows us to integrate a function over a three-dimensional space. Imagine finding the volume in 3D space or the mass of an object with varying density. These integrals give us such capabilities. When dealing with triple integrals, we often write it as $${\iiint }_{D}f(x,y,z)dV$$where $D$ is your region in space, $f(x,y,z)$ is the function you're integrating, and $dV$ represents a small volume element.

Triple integrals often use different coordinate systems, like Cartesian or spherical, to facilitate easier calculation depending on the symmetry and complexity of the region. It's essential to understand the space you're working with and choose the most effective coordinate system.

Jacobian Determinant

The Jacobian determinant is a critical component when performing coordinate transformations, such as for a change of variables in multiple integrals. It essentially helps us understand how the transformation skews or scales volume elements as we move from one coordinate system to another. This determinant comes from a matrix that contains all the partial derivatives of each new variable with respect to each old variable.

For example, if we have a transformation from variables $(x,y,z)$ to $(u,v,w)$, the Jacobian $J$ is given by:$$J=\left|\begin{array}{ccccccc}\frac{\partial x}{\partial u}& \frac{\partial x}{\partial v}& \frac{\partial x}{\partial w}\frac{\partial y}{\partial u}& \frac{\partial y}{\partial v}& \frac{\partial y}{\partial w}\frac{\partial z}{\partial u}& \frac{\partial z}{\partial v}& \frac{\partial z}{\partial w}\end{array}\right|$$A positive determinant indicates a volume-preserving transformation, whereas a zero or negative value can signal other transformations like reflections or collapses to a lower dimension.

Integration Bounds

Integration bounds define the limits of integration for each of the variables involved in the integral. These bounds determine the specific region over which the function is being integrated. In a triple integral, three such bounds indicate the domain or volume within the coordinate space.

For the change of variables, the original complex bounds are converted into simpler, often rectangular bounds, which drastically simplifies the integral. For instance, when using a transformation specified by:

• $u$ bounded by planes gives more manageable bounds in one dimension.
• $v$ and $w$ similarly transformed offer easier or standardized intervals like $0$ to $1$ or $0$ to $3$.

This conversion forms the heart of transforming an integral into a more calculable form.

Volume Integral

The volume integral is an application of the triple integral that calculates the total volume enclosed by given boundaries. When switching variables, it becomes essential to adjust both the equation and the limits to reflect this transformation.

Volume integrals are perfect for finding volumes in irregularly shaped regions, aided by transformations that simplify these shapes. In our example, integrating over $(u,v,w)$ space with transformed constant limits simplifies the integrals greatly.

The integral evaluates as:$${\iiint }_{D^{\prime }}|J|dudvdw$$This gives the volume of the region, scaled by the Jacobian. Essentially, understanding volume integrals allows us to explore and quantify three-dimensional spaces efficiently, especially when symmetry or pre-defined shapes do not apply.