Problem 9 - APEX Calculus

Question

(a) Evaluate the given iterated integral, and (b) rewrite the integral using the other order of integration. $$ \int_{0}^{1} \int_{-\sqrt{1-y}}^{\sqrt{1-y}}(x+y+2) d x d y $$

Step-by-Step Solution

Answer

The evaluated integral is \(\frac{16}{3}\), and with changed order, \(\int_{-1}^{1} \int_{0}^{1-x^2} (x+y+2) \ dy \ dx\).

Step 1: Evaluate the Inner Integral

The given iterated integral is \( \int_{0}^{1} \int_{-\sqrt{1-y}}^{\sqrt{1-y}}(x+y+2) d x d y \). Start by evaluating the inner integral with respect to \( x \): \( \int_{-\sqrt{1-y}}^{\sqrt{1-y}}(x+y+2) \ dx \). First, integrate the expression \((x+y+2)\) with respect to \(x\):\[\frac{x^2}{2} + yx + 2x\]between the limits \(-\sqrt{1-y}\) and \(\sqrt{1-y}\).

Step 2: Apply the Limits of the Inner Integral

Evaluate the antiderivative:\[\left[ \frac{x^2}{2} + yx + 2x \right]_{-\sqrt{1-y}}^{\sqrt{1-y}} = \left( \frac{(\sqrt{1-y})^2}{2} + y(\sqrt{1-y}) + 2(\sqrt{1-y}) \right) - \left( \frac{(-\sqrt{1-y})^2}{2} + y(-\sqrt{1-y}) + 2(-\sqrt{1-y}) \right)\]. Both terms in the square roots simplify symmetrically to \(\frac{1-y}{2} + y\sqrt{1-y} + 2\sqrt{1-y}\) and \(-\frac{1-y}{2} - y\sqrt{1-y} - 2\sqrt{1-y}\).

Step 3: Simplify Expressions

Simplify the expression derived from the previous step: \[\left( \frac{1-y}{2} + y\sqrt{1-y} + 2\sqrt{1-y} \right) + \left( \frac{1-y}{2} + y\sqrt{1-y} + 2\sqrt{1-y} \right)\]. This results in \[ (1-y) + 2y\sqrt{1-y} + 4\sqrt{1-y} \].

Step 4: Evaluate the Outer Integral

Now integrate \((1-y) + 2y\sqrt{1-y} + 4\sqrt{1-y}\) from \(y = 0\) to \(y = 1\). Simplifying further: this becomes \(2(\sqrt{1-y})(y+2)\). Compute \( \int_{0}^{1} \left((1-y) + 2(y + 2)\sqrt{1-y} \right) dy\).

Step 5: Change Order of Integration

To change the order of integration, first determine the limits for \(x\) and \(y\). The original region is defined by \(-\sqrt{1-y} \le x \le \sqrt{1-y}\) and \(0 \le y \le 1\). The "outer" range \(y=0\) to \(y=1\) corresponds to \(x=0\) to a boundary condition for \(y = 1 - x^2\). Write the integral as \[\int_{-1}^{1} \int_{0}^{1-x^2} (x+y+2) \ dy \ dx\].

Key Concepts

Order of Integration

When dealing with iterated integrals, the order in which you integrate can significantly impact the ease of computation. Typically, an iterated integral is expressed as either \[ \int_a^b \int_{c(y)}^{d(y)} f(x, y) \, dx \, dy \] or \[ \int_c^d \int_{a(x)}^{b(x)} f(x, y) \, dy \, dx \]. In the first format, you integrate with respect to \( x \) first, then \( y \). In the second, you reverse the order. This flexibility allows you to choose the order that simplifies calculations, by considering the function's form or the integration limits.Remember:

• The choice of the integration order can simplify or complicate the problem.
• Always consider how the function and its limits relate to each other.

Order matters and can lead to more straightforward results depending on the problem's nature.

Integration Limits

Integration limits define the region over which you perform integration. They are the boundaries that enclose the area or volume you're integrating over.In our original exercise, the limits were expressed as \( \int_{0}^{1} \int_{-\sqrt{1-y}}^{\sqrt{1-y}} (x+y+2) \, dx \, dy \). Here:

• \( y \) bounds range from 0 to 1.
• For each \( y \), \( x \) bounds range between \(-\sqrt{1-y}\) and \(\sqrt{1-y}\).

This creates a region defined by a half-circle in the \( xy \)-plane.When changing the order of integration, adjust the limits accordingly. The new limits must still cover the same region. For instance, after transforming the integral, observe that for each \( x \) (from -1 to 1), \( y \) is bounded from 0 to \(1 - x^2\). This confines the area to the same circular region defined earlier.

Changing Integration Order

Sometimes, it is advantageous to change the order of integration. This can simplify the computation depending on the bounds and the integrand's structure.To change the order, first visualize or sketch the region of integration. Identify the boundaries of the region to redefine them. Originally, we had:\[ \int_{0}^{1} \int_{-\sqrt{1-y}}^{\sqrt{1-y}} (x+y+2) \, dx \, dy \] This was re-expressed as:\[ \int_{-1}^{1} \int_{0}^{1-x^2} (x+y+2) \, dy \, dx \].

• The integrals' limits reflect a change that makes the vertical and horizontal slices easier to manage.
• Ensure the new limits cover the same region; visualize, then redefine.

Changing the integration order can be particularly useful with more complex regions or functions.

Evaluating Integrals

Evaluating iterated integrals requires careful handling of function expressions and their bounds. Start by solving the inner integral first. This usually involves finding an antiderivative with respect to one variable while treating others as constants. For example:\[ \int_{-\sqrt{1-y}}^{\sqrt{1-y}}(x+y+2) \, dx \] involves treating \( y \) as a constant and evaluating the result.After finding the antiderivative:

• Apply the integration limits to get a simplified expression.
• Proceed to integrate the resulting expression with respect to the next variable.

Continuing with the evaluation, use the principle \[ F(b) - F(a) \] to apply the limits effectively, acquiring your result.Each layer of integration brings you closer to a final numerical value that represents the sum over the specified region. Care and precision in each step ensure accuracy in the final result.