Problem 30 - University Calculus: Early Transcendentals

Question

Each gives an integral over a region in a Cartesian coordinate plane. Sketch the region and evaluate the integral. $$\int_{0}^{1} \int_{0}^{\sqrt{1-s^{2}}} 8 t d t d s \quad(\text { the } s t-\text { plane })$$

Step-by-Step Solution

Answer

The integral evaluates to \(\frac{8}{3}\).

Step 1: Understand the Integration Limits

The integral \( \int_{0}^{1} \int_{0}^{\sqrt{1-s^{2}}} 8t \, dt \, ds \) involves two variables, \(s\) and \(t\). The outer integral is from \(s=0\) to \(s=1\) while the inner integral is from \(t=0\) to \(t=\sqrt{1-s^2}\). This indicates that for each fixed \(s\), the variable \(t\) ranges from 0 to \(\sqrt{1-s^2}\), forming the upper bound of a semicircle of radius 1.

Step 2: Sketch the Region of Integration

In the \(st\)-plane, the region is bounded by \(t = 0\) (the x-axis), \(s = 0\), \(s = 1\), and \(t = \sqrt{1-s^2}\), which describes the upper half of the unit circle centered at the origin. It extends from \(s=0\) to \(s=1\), and \(t\) traces the semicircular arc from \(t=0\) to \(t=1\).

Step 3: Perform the Inner Integration

Evaluate the inner integral \( \int_{0}^{\sqrt{1-s^2}} 8t \, dt \). To find this, integrate 8t with respect to \(t\). Using the power rule for integration, we get: \(\int 8t \, dt = 8 \cdot \frac{t^2}{2} = 4t^2\). Following this, evaluate it from \(t = 0\) to \(t = \sqrt{1-s^2}\).

Step 4: Evaluate the Inner Integral

Substitute the limits into \(4t^2\): \[ 4(\sqrt{1-s^2})^2 - 4(0)^2 = 4(1-s^2). \]Simplifying gives \(4 - 4s^2\) as the result of the inner integration.

Step 5: Perform the Outer Integration

Now integrate the result of the inner integral with respect to \(s\): \[ \int_{0}^{1} (4 - 4s^2) \, ds. \]Split the integration: \(\int_{0}^{1} 4 \, ds - \int_{0}^{1} 4s^2 \, ds.\)

Step 6: Evaluate the Outer Integral

Calculate each part: \(\int_{0}^{1} 4 \, ds = 4s \Big|_{0}^{1} = 4(1) - 4(0) = 4\).For \(\int_{0}^{1} 4s^2 \, ds\), use the power rule: \( \int s^n \, ds = \frac{s^{n+1}}{n+1} \), giving \(\int s^2 \, ds = \frac{s^{3}}{3}.\)So, \(4 \int s^2 \, ds = \frac{4s^3}{3} \Big|_{0}^{1} = \frac{4}{3}\).

Step 7: Calculate the Final Result

Combine the results from both parts of the outer integral: \[ 4 - \frac{4}{3} = \frac{12}{3} - \frac{4}{3} = \frac{8}{3}. \]Thus, the value of the integral \( \int_{0}^{1} \int_{0}^{\sqrt{1-s^2}} 8t \, dt \, ds \) is \(\frac{8}{3}\).

Key Concepts

Double Integration

Double integration is a process used to compute the volume under a surface over a given region. In this exercise, the double integral \( \int_{0}^{1} \int_{0}^{\sqrt{1-s^{2}}} 8t \, dt \, ds \) allows us to measure the three-dimensional volume that lies beneath the surface formed by the function \(8t\) in the region described. There are two levels of integration involved:

• The inner integral \( \int_{0}^{\sqrt{1-s^2}} 8t \, dt \) calculates the area in the direction of the variable \(t\) for a fixed \(s\).
• The outer integral \( \int_{0}^{1} ( \cdot ) \, ds \) then integrates the slice obtained from the inner integral across the specified range for \(s\).

This layered integration approach allows for breaking down complex three-dimensional problems into manageable two-dimensional slices, evaluated step-by-step.

Polar Coordinates

In certain cases, converting Cartesian coordinates to polar coordinates can simplify the integration process. Polar coordinates \((r, \theta)\) utilize radius \(r\) from the origin and angle \(\theta\), offering a circular perspective instead of perpendicular axes (\(x, y\) or \(s, t\)). This exercise's formula \(t = \sqrt{1-s^2}\) resembles part of the polar coordinate equation for a circle, \(r^2 = x^2 + y^2\). Here, it represents a semicircle on the Cartesian plane but implies a circular region beneficial in polar representation. Switching to polar coordinates yields integrals often simpler to solve. The given semicircular boundary could be replaced by:

• \(s = r\cos(\theta)\)
• \(t = r\sin(\theta)\)

The transformation and new limits make revisiting double integration in polar form an excellent task for further practice and deeper comprehension.

Region of Integration

The region of integration is the area over which the double integral is evaluated. Understanding this concept is crucial because it defines where the calculation takes place. In the given exercise, the integration region is bounded by curves on the \(st\)-plane:

• From \(t=0\) (the x-axis) to \(t=\sqrt{1-s^2}\), tracing the upper semicircle.
• From \(s=0\) to \(s=1\), ensuring the limit spans correctly across the \(s\)-axis.

Graphically, this forms a semicircular region on the \([0, 1]\) interval. Visualizing these boundaries helps emphasize the slice-and-sum approach in double integration. Whether drawing or imagining the region, recognizing each boundary and its role in containment makes calculating area or volume more intuitive.