Problem 41 - University Calculus: Early Transcendentals

Question

Sketch the region of integration and write an equivalent double integral with the order of integration reversed. $$\int_{0}^{1} \int_{-\sqrt{1-y^{2}}}^{\sqrt{1-y^{2}}} 3 y d x d y$$

Step-by-Step Solution

Answer

Reversed integral is \(\int_{-1}^{1} \int_{0}^{\sqrt{1-x^{2}}} 3y \ dy \ dx\).

Step 1: Understand the Given Integral

We have the double integral \( \int_{0}^{1} \int_{-\sqrt{1-y^{2}}}^{\sqrt{1-y^{2}}} 3y \ dx \ dy \). This represents integration over a region in the \(xy\)-plane.

Step 2: Identify the Limits of Integration

The inner integral \( \int_{-\sqrt{1-y^{2}}}^{\sqrt{1-y^{2}}} dx \) indicates the range of \(x\) for each fixed \(y\), while \(y\) ranges from 0 to 1 in the outer integral.

Step 3: Sketch the Region of Integration

The limits \(x = -\sqrt{1-y^{2}}\) to \(x = \sqrt{1-y^{2}}\) describe a semi-circle with radius 1 on the \(xy\)-plane. Since \(y\) ranges from 0 to 1, draw the upper half of a circle centered at the origin \((0,0)\) extending up to \(y = 1\).

Step 4: Describe the Reversal of Integration

To reverse the order of integration, we first note that the curve \(x^2 + y^2 = 1\) represents the boundary of the region, and so the integration region is the upper semi-circle.

Step 5: Set New Integration Limits for \(x\)

For the reversed order, \(x\) ranges from \(-1\) to \(1\), which are the left and right boundaries of the semi-circle in the \(x\)-axis.

Step 6: Determine Limits for \(y\) Based on \(x\)

For each fixed \(x\), \(y\) ranges from 0 up to the semi-circle boundary, described by \(y = \sqrt{1-x^{2}}\), due to the equation \(x^2 + y^2 = 1\).

Step 7: Write the Reversed Integral

The reversed order of the integral becomes: \[ \int_{-1}^{1} \int_{0}^{\sqrt{1-x^{2}}} 3y \ dy \ dx \]

Key Concepts

Region of Integration

When dealing with double integrals, the **region of integration** is crucial as it defines the part of the plane where the integration occurs. In this problem, the original double integral \[ \int_{0}^{1} \int_{-\sqrt{1-y^{2}}}^{\sqrt{1-y^{2}}} 3y \, dx \, dy \]shows us that the region of integration is surface-level defined by the limits in terms of \(x\) and \(y\).
Here, the inner integral limits \( -\sqrt{1-y^{2}} \leq x \leq \sqrt{1-y^{2}} \) suggest that for each \(y\) value ranging from 0 to 1 (as per the outer integral), \(x\) values are confined within a semicircular boundary. Together with \(y\) ranging up to 1, this forms a semi-circle on the upper half of the \(xy\)-plane with a radius of 1 centered at the origin.
Recognizing this region is key. It helps ensure that our calculations relate to the correct area on the plane. Visualizing the area as the upper half of a circle simplifies understanding of the integration limits and helps with any adjustments necessary, such as changing the order of integration.

Order of Integration

The **order of integration** refers to the sequence in which we integrate multiple variables. In our initial integral \[ \int_{0}^{1} \int_{-\sqrt{1-y^{2}}}^{\sqrt{1-y^{2}}} 3y \, dx \, dy \]the integration is performed first with respect to \(x\) and then \(y\). This order dictates how we navigate through the region of integration.
The given order tells us to consider horizontal slices of the region, moving vertically from \(y = 0\) to \(y = 1\). For each fixed value of \(y\), \(x\) varies between the limits defined by \(-\sqrt{1-y^{2}}\) to \(\sqrt{1-y^{2}}\), depicting how wide the slices extend horizontally. Reflecting on this order helps in setting up the integration and ensures the integral covers the exact intended area.

• Inner integral: dealt with \(x\), determining slice width.
• Outer integral: dealt with \(y\), determining overall slice direction.

Changing the order can sometimes simplify the math, depending on the region's symmetry or boundaries.

Double Integral

Double integrals enable us to calculate the volume under a surface over a certain region in the plane. In our example, the integral \[ \int_{0}^{1} \int_{-\sqrt{1-y^{2}}}^{\sqrt{1-y^{2}}} 3y \, dx \, dy \]is a double integral that calculates such a volume for the expression \(3y\) over the specified semi-circular area.
The process involves:

• First integrating with respect to one variable (\(x\) in the original order), which narrows down a 'partial volume' above slices parallel to the \(y\)-axis.
• Then integrating the result with respect to the other variable (\(y\)), compiling these partial volumes over the entire region.
  

Double integrals are powerful tools in applications like physics and engineering, where they can model complex physical properties over specified regions, such as mass, charge, or probability density. Understanding their geometric interpretation helps in devising strategies for solving them effectively.

Reversal of Integration Order

Sometimes reversing the **order of integration** can make the computation easier. This involves swapping the roles of the inner and outer integration.
In our case, changing from\[ \int_{0}^{1} \int_{-\sqrt{1-y^{2}}}^{\sqrt{1-y^{2}}} 3y \, dx \, dy \]to\[ \int_{-1}^{1} \int_{0}^{\sqrt{1-x^{2}}} 3y \, dy \, dx \]means reversing the process by integrating first with respect to \(y\), then \(x\).
Steps involved in reversal include:

• Identifying the new outer integration limits: \(-1 \leq x \leq 1\) (spanning the horizontal diameter of the circle).
• Determining new inner limits of \(y\), which are from the horizontal line \(y = 0\) and up to the semi-circle \(y = \sqrt{1-x^{2}}\).

The reversal simplifies handling the integral by tailoring the limits to the region's shape, often leading to simpler or more intuitive substitutions and integration paths.