Question

Evaluate the following integrals using a change of variables of your choice. Sketch the original and new regions of integration, \(R\) and \(S\). $$\int_{0}^{1} \int_{y}^{y+2} \sqrt{x-y} d x d y$$

Short answer

Question: Evaluate the integral $$\int_{0}^{1} \int_{y}^{y+2} \sqrt{x-y} d x d y$$. Answer: The value of the integral is \(\frac{4\sqrt{2}}{3}\).

Step-by-step solution

Find the region R in the x-y plane

Observe the limits of integration. The limits for \(y\) are from 0 to 1. The limits for \(x\) are determined by the limits for \(y\) which are from \(y\) to \(y+2\). Now, let's find the corner points of the region R, where the boundaries of the integral intersect: 1. \((0,0)\): lower limit of y and lower limit of x (when \(y=0\), \(x=0\)), 2. \((2,0)\): lower limit of y and upper limit of x (when \(y=0\), \(x=y+2=2\)), 3. \((3,1)\): upper limit of y and upper limit of x (when \(y=1\), \(x=y+2=3\)), 4. \((1,1)\): upper limit of y and lower limit of x (when \(y=1\), \(x=1\)). These corner points form a trapezoid in the x-y plane.

Choose a suitable change of variables for transformation

We will choose the variable transformation: $$u = x - y \\ v = y$$ This transformation eliminates the dependence of x limits on y, and the region S in the u-v plane will be rectangular. The inverse transformation is: $$x = u + v \\ y = v$$ Now, we will find the corner points of the new region S by applying the transformation to the corner points of region R. 1. \((0,0)\): \((u,v) = (0,0)\) 2. \((2,0)\): \((u,v) = (2,0)\) 3. \((3,1)\): \((u,v) = (2,1)\) 4. \((1,1)\): \((u,v) = (0,1)\) After the transformation, the new region S is a rectangle with vertices \((0,0), (2,0), (2,1),\) and \((0,1)\).

Find the Jacobian

Now, let's find the Jacobian of the transformation. $$J(u,v) = \frac{\partial(x, y)}{\partial(u, v)} = \begin{vmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{vmatrix} = \begin{vmatrix} 1 & 1 \\ 0 & 1 \end{vmatrix} = (1)(1) - (0)(1) = 1$$ Since the Jacobian is nonzero, the transformation is valid.

Rewrite the integral in terms of u and v

Now, we rewrite the integrand in terms of u and v using the inverse transformation: $$\sqrt{x-y} = \sqrt{(u+v)-v} = \sqrt{u}$$ Next, we replace the limits of integration with the limits in u-v plane: $$\int_{0}^{1} \int_{y}^{y+2} \sqrt{x-y} dx dy = \int_{0}^{1} \int_{0}^{2} \sqrt{u} |J(u,v)| du dv = \int_{0}^{1} \int_{0}^{2} \sqrt{u} du dv$$

Evaluate the transformed integral

Finally, we evaluate the integral: $$\int_{0}^{1} \int_{0}^{2} \sqrt{u} du dv = \int_{0}^{1} \left[\frac{2}{3}u^{3/2}\right]_0^2 dv = \int_{0}^{1} \left(\frac{2}{3}(2^{3/2})\right) dv = \frac{4\sqrt{2}}{3} \int_{0}^{1} dv = \frac{4\sqrt{2}}{3} [v]_0^1 = \boxed{\frac{4\sqrt{2}}{3}}$$ Thus, the value of the integral is \(\frac{4\sqrt{2}}{3}\).

Key concepts

Change of Variables

The change of variables is a common technique in integration that simplifies the process by transforming it into a more straightforward form. In the given problem, the transformation

• \( u = x - y \)
• \( v = y \)

allows us to evaluate the integral by transforming the region of integration and the integrand simultaneously. This transformation effectively decouples the x-dependence of the limits of integration, making it easier to compute the integral. When choosing a new set of variables, it's important to consider how they might simplify both the integrand and the bounds of the region of integration. This method is often compared to substitution in single-variable integration, but it is applied to multiple variables.

Multiple Integrals

Multiple integrals extend the concept of integration to functions with more than one variable. They are used to calculate volume under surfaces or integrate over regions in multiple dimensions. In our problem, we deal with double integrals over a region defined in the xy-plane. The process involves:

• Setting up the integral with respect to one variable, while treating the other as a constant.
• Integrating in sequence, often referred to as iterated integration.

The integration bounds are determined by the problem's region of integration, which can be complicated by dependencies between variables. Here, the change of variables simplifies this by transforming the trapezoidal region R into a rectangular region S, with straightforward limits \(0 \leq u \leq 2\) and \(0 \leq v \leq 1\). This transformation makes evaluating the integral more manageable.

Jacobian Determinant

The Jacobian determinant is a critical factor in multivariable transformations, particularly in changing the variables in integration. It measures the

• Rate of change of area (or volume in higher dimensions) during the transformation process.

For our variable transformation, the Jacobian \( J(u, v) \) is calculated from the partial derivatives of the original variables (x, y) with respect to the new variables (u, v). In this case, we find:\[J(u, v) = \begin{vmatrix}\frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\frac{\partial y}{\partial u} & \frac{\partial y}{\partial v}\end{vmatrix} = \begin{vmatrix}1 & 1 \0 & 1\end{vmatrix} = 1\]The non-zero Jacobian validates the transformation, indicating that the coordinate system change properly represents area, preserving the essence of the integral.

Region of Integration

Defining the region of integration is fundamental in setting multiple integrals. The region for our initial integral is represented as a trapezoid in the xy-plane, determined by the limits on x and y. Corner points at

• \((0,0)\)
• \((2,0)\)
• \((3,1)\)
• \((1,1)\)

describe the bounds of this region. Upon changing variables, the region becomes a rectangle in the uv-plane. The new boundaries are easier to handle

• from \(u = 0\) to \(u = 2\)
• and \(v = 0\) to \(v = 1\)

This simplification transforms the problem from iterating over a complexly bounded shape to one with uniform width and height, which is much simpler to integrate over. Identifying the region correctly is crucial, as it directly impacts the accuracy and ease of solving the integral.