Problem 11 - Advanced Engineering Mathematics

Question

Sketch the region of integration for the given iterated integral. $$ \int_{-1}^{3} \int_{0}^{\sqrt{16-y^{2}}} f(x, y) d x d y $$

Step-by-Step Solution

Answer

The region is the right half of the circle \( x^2 + y^2 = 16 \) between \( y = -1 \) and \( y = 3 \).

Step 1: Understand the Limits of Integration

The given iterated integral is \( \int_{-1}^{3} \int_{0}^{\sqrt{16-y^2}} f(x, y) \, dx \, dy \). The outer integral \( \int_{-1}^{3} \) indicates that \( y \) varies from \(-1\) to \(3\). The inner integral \( \int_{0}^{\sqrt{16-y^2}} \) shows that for a fixed \( y \), \( x \) varies from \(0\) to \(\sqrt{16-y^2}\).

Step 2: Analyze the Inner Integral

The expression \( x = \sqrt{16 - y^2} \) describes the boundary of the region in the \( xy \)-plane. Squaring both sides, we have \( x^2 = 16 - y^2 \), which rearranges to \( x^2 + y^2 = 16 \). This represents a circle centered at \((0, 0)\) with radius \(4\). However, because \( x \geq 0 \), we are only interested in the right half of this circle.

Step 3: Define the Region of \( y \)

With \( y \) ranging from \(-1\) to \(3\), the region of interest is part of the circle \( x^2 + y^2 = 16 \) within these values. Since \( y \) continues through the range \(-1\) to \(3\), it implies that the upper portion stretches from these values.

Step 4: Sketch the Region of Integration

In the \( xy \)-plane, sketch the circle \( x^2 + y^2 = 16 \) which goes through \((4, 0)\), \((0, 4)\), \((-4, 0)\) and \((0, -4)\). Because \( x \) is non-negative, only the right semicircle (including the y-axis as a boundary) is considered. Overlay this with horizontal boundaries at \( y = -1 \) and \( y = 3 \). The region of integration is thus the right half of the circle from \( y = -1 \) to \( y = 3 \).

Key Concepts

Limits of Integration

When you're dealing with an iterated integral like the one provided in the exercise, understanding the limits of integration is crucial.

• The outer integral, which goes from \(-1\) to \(+3\), sets the range for the variable \(y\). So, \(y\) can be any value between these two numbers.


• The inner integral dictates the range for \(x\) and it varies from \(0\) to \(\sqrt{16-y^2}\). This means \(x\) will adjust according to the current value of \(y\).

These limits help define the specific region on the \(xy\)-plane where the function \(f(x, y)\) needs to be evaluated. The key here is that the limits are not constant for \(x\), as they depend on \(y\). So as \(y\) moves between \(-1\) and \(+3\), \(x\) must be recalculated accordingly, using the formula inside the square root.

Polar Coordinates

Polar coordinates offer a different way to describe points in a plane, using angles and distances instead of the traditional xy-coordinates. They can often simplify problems involving circular regions.

• In polar coordinates, any point in the plane is defined by its distance \(r\) from the origin and the angle \(\theta\) it makes with the positive x-axis.


• In our exercise, we are already dealing with a circular boundary given by \((x^2 + y^2 = 16)\). This is naturally described in polar coordinates, as \(r = 4\).

However, the exercise is set up in Cartesian coordinates, which is fine given the problem's requirements. If the task were to convert these integrals into polar, you'd need functions of \(r\) and \(\theta\) instead, which can sometimes simplify integration of functions with circular symmetry.

Circle Equation

The equation \(x^2 + y^2 = 16\) represents a circle on the xy-plane. Here's the breakdown:

• This specific equation describes a circle centered at the origin \(0,0\).


• The number \(16\) is the square of the radius of the circle, so the radius is \(4\).

Understanding this allows you to visualize the exact boundary that the iterated integral outlines. Every point \(x, y\) on the circle is at a distance \(4\) from the center. The exercise focuses on the right half of this circle because \(x\) must remain non-negative.

Semicircle

In this task, you need to identify that the region of integration involves a semicircle. Since \(x\) is constrained to positive values:

• We focus only on the right half of the full circle \(x^2 + y^2 = 16\).
• This right semicircle is bounded vertically by the lines \(y = -1\) and \(y = 3\), meaning it doesn't stretch from \(-4\) to \(+4\) as a full circle would.

The region gets a bit more defined here as you're considering points where \(0 \leq x \leq \sqrt{16-y^2}\) and \(-1 \leq y \leq 3\), helping to isolate only a portion of the semicircle. This sketch effectively captures this nuanced section of the circle, getting only the needed part for evaluating the integral.