Problem 7 - Thomas Calculus

Question

Sketch the described regions of integration. \begin{equation}0 \leq y \leq 1, \quad 0 \leq x \leq \sin ^{-1} y\end{equation}

Step-by-Step Solution

Answer

The region is between \(x = 0\) and \(x = \sin^{-1}(y)\), for \(y\) from 0 to 1.

Step 1: Understand Region Boundaries

The region is defined by two inequalities: \( 0 \leq y \leq 1 \) and \( 0 \leq x \leq \sin^{-1}(y) \). This means \( y \) ranges between 0 and 1, and for each \( y \), \( x \) ranges from 0 to \( \sin^{-1}(y) \).

Step 2: Interpret Inverse Sine Function

Recall that \( \sin^{-1}(y) \) represents the angle whose sine is \( y \). The function \( \sin^{-1}(y) \) is defined for \( y \) values between -1 and 1, but here we only consider \( y \) between 0 and 1, which corresponds to angles from 0 to \( \pi/2 \).

Step 3: Create Coordinate Transformation for Clarity

To make the region clearer, consider the transformation \( x = \sin^{-1}(y) \). This helps us understand that for every \( y \), \( x \) goes from 0 (the y-axis) up to the angle whose sine is \( y \).

Step 4: Sketch the Region

On a 2D coordinate system, draw the line \( y = 0 \) and \( y = 1 \). For each value in between, the x-coordinate goes from 0 to \( \sin^{-1}(y) \). This creates a curved line corresponding to \( x = \sin^{-1}(y) \), which sweeps from \( (0,0) \) progressively to \( (\pi/2,1) \).

Step 5: Identify the Shape of Region

The region is typically bounded on the left by the y-axis \(x=0\), horizontally from 0 to \(\sin^{-1}(y)\), mapping a shape bounded between the line \(x = \sin^{-1}(y)\) and the y-axis from bottom \(y = 0\) to top \(y = 1\).

Key Concepts

Regions of Integration

In calculus, understanding regions of integration is crucial when solving double integrals. These regions help define where integration takes place. For the given problem, the integration occurs in a specific area on the coordinate plane. This area, or region, is bounded by specific equations or inequalities.

Here's how to interpret the given region:

• The inequality \(0 \leq y \leq 1\) indicates that the region spans vertically from \(y = 0\) to \(y = 1\).
• The other condition, \(0 \leq x \leq \sin^{-1}(y)\), means that for each fixed \(y\), the value of \(x\) ranges from the y-axis up to a curved line defined by the inverse sine function. This plots out as a vertical strip for each value of \(y\).

The complete region comprises all points (\(x, y\)) that fit within both inequalities. This produces a curved triangular-like region below the curve \(x = \sin^{-1}(y)\) in the first quadrant.

Inverse Functions

Inverse functions, like \(\sin^{-1}(y)\), play a pivotal role in transforming and understanding functions by reversing their outputs and inputs. Specifically, \(\sin^{-1}(y)\) is the inverse sine function, and it enables us to determine the angle whose sine equals \(y\).

Important points about inverse functions:

• The domain for \(\sin^{-1}(y)\) is \([-1, 1]\), but our problem focuses on \([0, 1]\), giving values from 0 to \(\pi/2\).
• In our integration exercise, understanding \(\sin^{-1}(y)\) helps determine the bounds for \(x\). It tells us the maximum \(x\) for a given \(y\) by calculating the necessary angle.

These functions help identify specific points and regions where particular conditions on the range, like the maximum width of rectangles forming a region, occur.

Coordinate Transformation

Coordinate transformations make it easier to understand complex geometric regions by changing how we look at them. In our exercise, the transformation through \(x = \sin^{-1}(y)\) helps clarify our region of integration.

Why use transformations:

• They can simplify complex regions into more understandable forms. For instance, using \(x = \sin^{-1}(y)\) helps visualize how \(x\) varies with \(y\).
• Transformations can align regions with new axes, making calculations of area, volume, or other properties more straightforward.

This approach can make regions that are tricky to envision in normal coordinates easier to map out, interpret, and utilize in calculations or further analysis.

Integration Bounds

Determining integration bounds is essential in setting up integrals. These bounds define where the integration starts and stops along each axis and are influenced by the functions that mold the region.

In this specific problem:

• The bounds for \(y\) were straightforward: from 0 to 1.
• For \(x\), the bounds are described by \(0\) to \(\sin^{-1}(y)\). This complexity arises from \(x\) not being constant but instead becoming a function of \(y\).

Understanding these bounds allows for integrating over an area where the conditions of the problem hold true. Properly established integration limits are fundamental in ensuring calculations accurately represent the desired region.