Problem 32 - Thomas Calculus

Question

Gives an integral over a region in a Cartesian coordinate plane. Sketch the region and evaluate the integral. \begin{equation} \int_{0}^{3 / 2} \int_{1}^{4-2 u} \frac{4-2 u}{v^{2}} d v d u \quad(\text { the } u v \text { -plane }) \end{equation}

Step-by-Step Solution

Answer

The value of the integral is \( \frac{9}{4} \).

Step 1: Understand the Region of Integration

The integral is given in the form \( \int_{a}^{b} \int_{c(u)}^{d(u)} f(v, u) \, dv \, du \). It involves two nested integrals, with an outer integral in terms of \( u \) from 0 to \( \frac{3}{2} \), and an inner integral in terms of \( v \) from 1 to \( 4 - 2u \). The region of integration corresponds to these bounds in the \( uv \)-plane.

Step 2: Sketch the Region

To sketch the region on the \( u v \)-plane, consider the constraints: \( 0 \leq u \leq \frac{3}{2} \) and \( 1 \leq v \leq 4 - 2u \). The line \( v = 4 - 2u \) intersects the \( v \)-axis at \( v = 4 \) and the \( u \)-axis at \( u = 2 \). The region lies under this line, is right of \( v = 1 \) and above \( u = 0 \). At \( u = \frac{3}{2} \), the line \( v = 4 - 2u \) gives \( v = 1 \), forming a triangle.

Step 3: Evaluate the Inner Integral

Evaluate \( \int_1^{4-2u} \frac{4-2u}{v^2} \, dv \). Set up the antiderivative: \( \int \frac{C}{v^2} \, dv = -\frac{C}{v} + C' \). Here, \( C = 4 - 2u \). Thus, \( \int \frac{4-2u}{v^2} \, dv = -\frac{4-2u}{v} \). Evaluate from 1 to \( 4-2u \): \( -\frac{4-2u}{4-2u} + \frac{4-2u}{1} = -1 + (4-2u) = 3 - 2u \).

Step 4: Evaluate the Outer Integral

Substitute the result of the inner integral into the outer integral: \( \int_0^{\frac{3}{2}} (3 - 2u) \, du \). Find the antiderivative \( 3u - u^2 \). Evaluate from 0 to \( \frac{3}{2} \): \( 3\frac{3}{2} - (\frac{3}{2})^2 = \frac{9}{2} - \frac{9}{4} = \frac{18}{4} - \frac{9}{4} = \frac{9}{4} \).

Step 5: Simplify the Final Result

The evaluation of the outer integral gives \( \frac{9}{4} \), which cannot be simplified further. Thus, this is the final answer.

Key Concepts

Region of Integration

When solving a double integral, the region of integration is crucial. It defines the limits within which the function is evaluated. In our example, the region is specified by the bounds of the integral:

• The outer integral extends from \( u = 0 \) to \( u = \frac{3}{2} \).
• The inner integral covers \( v \) from \( v = 1 \) to \( v = 4-2u \).

To understand it visually, sketch the region in the \( uv \)-plane:- The line \( v = 1 \) forms the lower horizontal boundary.- The line \( v = 4-2u \) is a downward sloping line that forms the upper boundary. This line intersects the \( v \)-axis at \( v = 4 \) and the \( u \)-axis at \( u = 2 \).- The region is then bounded on the left by \( u = 0 \), and its highest point intersects at \( u = \frac{3}{2} \) where \( v = 1 \). The outlined segment forms a triangular region, which needs to be considered during integration.

UV-Plane

The \( uv \)-plane is the coordinate system in which our problem is set. It is vital to visualize the integration process.

• The axes represent the variables \( u \) and \( v \).
• Understanding the geometry in this plane helps in evaluating the integral accurately.

The transformation from Cartesian coordinates to the \( uv \)-plane can simplify some calculations, allowing us to redefine the limits, and in some cases, the function form. Being comfortable with this transformation aids in recognizing simpler patterns and visualizing how the function behaves over specified intervals.

Nested Integrals

Nested or iterated integrals involve evaluating integrals within integrals. The given double integral \( \int_{0}^{3/2} \int_{1}^{4-2u} \frac{4-2u}{v^2} dv \, du \) illustrates this.

• The inner integral \( \int_{1}^{4-2u} \frac{4-2u}{v^2} \, dv \) is solved first with respect to \( v \).
• The result is then used in the outer integral \( \int_{0}^{3/2} (3-2u) \, du \) to find the final value.

Nesting simplifies complex multivariable integration into simpler single-variable integrals. Each step relies on finding the antiderivative and evaluating it over specified bounds. Careful tracking of each step ensures accuracy in the final computation.

Antiderivative Calculation

Calculating the antiderivative is a fundamental part of solving integrals. It involves reversing the differentiation process to find a function whose derivative is the integrand.For the inner integral \( \int \frac{4-2u}{v^2} dv \), we consider the antiderivative of \( \frac{C}{v^2} \),- Which results in \( -\frac{C}{v} + C' \) for a constant \( C \).- Here, \( C = 4-2u \), making the antiderivative \( -\frac{4-2u}{v} \).Once the inner integral is evaluated, the result is used in the outer integral to find its antiderivative, which is \( 3u - u^2 \). The limits of integration are then applied to compute the value. Developing a strong grasp of antiderivative calculation is key in tackling complex integrals effectively.