Problem 14 - Thomas Calculus

Question

Write an iterated integral for \(\iint_{R} d A\) over the described region \(R\) using (a) vertical cross-sections, (b) horizontal cross-sections. \begin{equation}\begin{array}{l}{\text { Bounded by } y=\tan x, x=0, \text { and } y=1} \end{array} \end{equation}

Step-by-Step Solution

Answer

The integrals are \( \int_{0}^{1} \int_{0}^{\tan^{-1}(y)} \, dx \, dy \) for vertical and \( \int_{0}^{\frac{\pi}{4}} \int_{\tan(x)}^{1} \, dy \, dx \) for horizontal cross-sections.

Step 1: Understand the Region R

The region \( R \) is bounded by \( y = \tan x \), \( x = 0 \), and \( y = 1 \). Visually, think of the graph of \( y = \tan x \) starting from \( x = 0 \) and identify where it reaches \( y = 1 \). This will help define the limits of integration.

Step 2: Vertical Cross-sections for Integration

For vertical cross-sections, with limits along the \( y \)-axis, the outer integral will be along \( y \) and the inner integral along \( x \). The region is bounded on the left by \( x = 0 \) and on the right by \( x = \tan^{-1}(y) \). Thus, the iterated integral is: \[\int_{0}^{1} \int_{0}^{\tan^{-1}(y)} \, dx \, dy\] Here the outer integral runs from \( y = 0 \) to \( y = 1 \).

Step 3: Horizontal Cross-sections for Integration

For horizontal cross-sections, with limits along the \( x \)-axis, the outer integral will be along \( x \) and the inner integral along \( y \). The region is bounded below by \( y = \tan(x) \) and above by \( y = 1 \). Thus, the iterated integral is: \[\int_{0}^{\frac{\pi}{4}} \int_{\tan(x)}^{1} \, dy \, dx\] Here, the outer integral runs from \( x = 0 \) to \( x = \frac{\pi}{4} \), as \( \tan(x) = 1 \) at \( x = \frac{\pi}{4} \).

Key Concepts

Bounded Regions

When trying to solve problems involving integrals over regions, understanding the bounded region is crucial. In this exercise, we're focusing on a region defined in the xy-plane. The boundaries are dictated by the equations \( y = \tan x \), \( x = 0 \), and \( y = 1 \). Let us break these down:

• The curve \( y = \tan x \) gives one boundary, representing a portion of the tangent function's graph.
• The line \( x = 0 \) corresponds to the y-axis, acting as the left boundary of the region.
• The horizontal line \( y = 1 \), indicates the upper limit of the region.

To visualize this region:\- Start from \( x = 0 \) and see where the function \( y = \tan x \) reaches \( y = 1 \).\- This happens at \( x = \frac{\pi}{4} \), since \( \tan(\frac{\pi}{4}) = 1 \).\The bounded region, therefore, is where all three conditions are met: the zone above the x-axis, below \( y = 1 \), and between \( x = 0 \) and \( x = \frac{\pi}{4} \). Understanding this helps set up the limits for both vertical and horizontal cross-sections.

Vertical Cross-sections

Vertical cross-sections involve slicing the region parallel to the y-axis. In simpler terms, imagine cutting the region up and down. With vertical cross-sections, we establish the outer integral along the y-direction, while the inner integral is along the x-direction.For our specific region:

• The outer integral bounds y from 0 to 1.
• With each fixed \( y \), the inner integral uses x, stretching from the left boundary at \( x = 0 \) to the curve that bounds on the right, given by \( x = \tan^{-1}(y) \).

This means, for each value of y between 0 and 1:\- We slice vertically, starting from the y-axis at \( x = 0 \), moving towards \( x = \tan^{-1}(y) \) where the curve \( y = \tan x \) sits.\Therefore, the iterated integral is:\[\int_{0}^{1} \int_{0}^{\tan^{-1}(y)} \, dx \, dy\]With this setup, the region is dissected into vertical strips, each corresponding to a y-value integration.

Horizontal Cross-sections

Horizontal cross-sections slice the region parallel to the x-axis. Picture cutting the region side to side. Here, we use x for the outer integral, with y described in the inner integral.For the described region:

• The outer integral bounds x from 0 to \( \frac{\pi}{4} \).
• For each have fixed \( x \), the inner integral's y ranges from the tangent curve \( y = \tan x \) to the line \( y = 1 \).

Thus, for each x within 0 to \( \frac{\pi}{4} \):\- We cut horizontally, starting below the tangent line at \( y = \tan(x) \) and going up to the constant line \( y = 1 \).\The iterated integral becomes:\[\int_{0}^{\frac{\pi}{4}} \int_{\tan(x)}^{1} \, dy \, dx\]With this approach, the region is sliced into horizontal layers, covering all possible y-values between \( \tan(x) \) and 1 for each slice.