Question

Use a change of variables to evaluate the following integrals. $$\begin{aligned} &\iiint_{D} x y d V ; D \text { is bounded by the planes } y-x=0\\\ &y-x=2, z-y=0, z-y=1, z=0, \text { and } z=3 \end{aligned}$$

Short answer

Answer: The value of the triple integral $\iiint_{D} x y dV$ is $\frac{8}{3}$.

Step-by-step solution

Identify the new coordinates

To simplify the integral, we will make a transformation of coordinates that aligns our new axes with the given planes. We can express the new coordinates as follows: $$\begin{aligned} u &= y-x\\ v &= y-x-2\\ w &= z-y \end{aligned}$$

Determine the Jacobian of the transformation

Now we must find the Jacobian of the transformation, which is given by the determinant of the matrix of partial derivatives. We first take the partial derivatives of the original coordinates with respect to the transformed coordinates: $$\begin{aligned} x &= \frac{1}{2}(2y -u -v)\\ y &= \frac{1}{2}(u+v)\\ z &= y+w \end{aligned}$$ The Jacobian matrix is given by: $$J=\begin{bmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} & \frac{\partial x}{\partial w}\\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} & \frac{\partial y}{\partial w}\\ \frac{\partial z}{\partial u} & \frac{\partial z}{\partial v} & \frac{\partial z}{\partial w} \end{bmatrix} = \begin{bmatrix} -\frac{1}{2} & -\frac{1}{2} & 0\\ \frac{1}{2} & \frac{1}{2} & 0\\ 0 & 0 & 1\end{bmatrix}$$ Now we compute the determinant of the Jacobian matrix: $$|J| = -\frac{1}{2}\begin{vmatrix} \frac{1}{2} & 0\\ 0 & 1\end{vmatrix} + \frac{1}{2}\begin{vmatrix} \frac{1}{2} & 0\\ 0 & 1\end{vmatrix} = -\frac{1}{2}\cdot\frac{1}{2}(1) + \frac{1}{2}\cdot\frac{1}{2}(1) = \frac{1}{2}$$

Determine the transformed region of integration

We now need to find the transformed region under the planes in the new coordinate space. We can do this by finding the bounds on the new variables, u, v, and w: $$\begin{aligned} 0 &= y-x &\Rightarrow{}& u = 0\\ 2 &= y-x &\Rightarrow{}& v = 0\\ 0 &= z-y &\Rightarrow{}& w = 0\\ 1 &= z-y &\Rightarrow{}& w = 1\\ 0 &= z &\Rightarrow{}& y+w = 0\\ 3 &= z &\Rightarrow{}& y+w = 3 \end{aligned}$$ Now we have the bounds on u, v, and w in our new coordinate system: $$\begin{aligned} D':&\\ 0 \leq u &\leq v\\ 0 \leq v &\leq 2\\ 0 \leq w &\leq 1\\ 0 \leq y+w &\leq 3 \end{aligned}$$

Transform and solve the triple integral

We can now transform our triple integral using the new coordinates: $$\begin{aligned} \iiint_{D} x y dV &= \iiint_{D'} \left(\frac{1}{2}(2y - u - v)\right)\left(\frac{1}{2}(u+v)\right)\left(\frac{1}{2}\, du \, dv \, dw\right)\\ &= \int_0^2 \int_0^v \int_0^1 \left(\frac{1}{2}(2y - u - v)\right)\left(\frac{1}{2}(u+v)\right)\left(\frac{1}{2}\right) du \, dv \, dw \end{aligned}$$ Now we solve the integral: $$\begin{aligned} \int_0^2 \int_0^v \int_0^1\left(\frac{1}{8}(2y - u - v)(u+v)\right) du \, dv \, dw &= \int_0^2 \int_0^v \left[\frac{1}{16}(u^2 + v^2 - 2uv)\right]_0^1 dv \, dw\\ &= \int_0^2 \int_0^v \frac{1}{16}(1 - 2v + v^2) dv \, dw\\ &= \int_0^2 \left[\frac{1}{48}(v^3 - 3v^2 + 3v)\right]_0^v dw\\ &= \int_0^2 \frac{1}{48}(8w^3 - 24w^2 + 24w) dw\\ &= \left[\frac{1}{192}(16w^4 - 64w^3 + 96w^2)\right]_0^2\\ &= \frac{1}{192}(256 - 512 + 768)\\ &= \frac{1}{192}(512) \end{aligned}$$ Thus, the value of the integral is: $$\iiint_{D} x y dV = \frac{1}{192}(512) = \boxed{\frac{8}{3}}.$$

Key concepts

Change of Variables

The change of variables technique is a crucial tool in evaluating complex integrals because it simplifies the form and limits of these integrals. This process involves finding new coordinates that make the region of integration simpler. In this exercise, we look for a coordinate transformation that aligns with the planes bounding the region.

When dealing with multivariable integrals, especially triple integrals, choosing the right set of new variables can make a daunting problem more manageable. Here, variables are selected so the bounds of integration are either constants or simpler linear functions.

• Choose new variables based on the bounds and symmetries.
• A beneficial change can render the integral into a more standard form.

This transformation may not only simplify the region but can also make the function itself simpler to integrate.

Jacobian Determinant

The Jacobian determinant plays a pivotal role in change of variables for multivariable integrals. It acts as the scaling factor for volume when transitioning between coordinate systems. The determinant of the Jacobian matrix gives the factor by which the integral must be adjusted to account for this change.

To compute the Jacobian, construct a matrix from the partial derivatives of the new coordinates in terms of the old ones. The resulting determinant composes the adjustment needed for the volume element.

• Matrix elements: partial derivatives of new variables with respect to old ones.
• The magnitude of the determinant: scale factor for the change of volume.

A correct Jacobian ensures that the transformed integral retains the same value as the original.

Coordinate Transformation

Coordinate transformation allows us to reframe the problem, making the integral easier to work with. It involves adopting a new perspective by aligning variables with the parameters that define the region's boundaries. This redefinition offers a clearer and often more straightforward path to solving the integral.

In our example, the chosen new coordinates, such as \(u = y-x\), were derived based on the given equation of planes, which helped in establishing simpler bounds.

• New coordinates often align with significant features of the problem (like planes).
• Convert to new variables that reflect natural limits and symmetries.

This proactive step ensures that we focus directly on the critical transformations needed for ease.

Integral Bounds

Determining the integral bounds in the new coordinate system is crucial for solving the integral correctly. These bounds define the region over which you are integrating, and a simplified set of bounds can make the evaluation process much smoother.

It involves rewriting the conditions set by the original boundaries into the language of the new variables. For example, in the transformation given in this exercise, boundaries like \(y-x=0\) become \(u=0\).

• Convert all original conditions into the language of new variables.
• Ensure bounds are consistent and correctly reflect the original integral domain.

The clearer the bounds are, the simpler the integration process will be, as seen when changing from complex boundaries to simple linear inequalities.