Problem 74 - Calculus Early Transcendentals

Question

Without evaluating integrals, explain why the following equalities are true. (Hint: Draw pictures.) a. \(\pi \int_{0}^{4}(8-2 x)^{2} d x=2 \pi \int_{0}^{8} y\left(4-\frac{y}{2}\right) d y\) b. \(\int_{0}^{2}\left(25-\left(x^{2}+1\right)^{2}\right) d x=2 \int_{1}^{5} y \sqrt{y-1} d y\)

Step-by-Step Solution

Answer

Question: Explain why the following equalities are true without actually evaluating the integrals: (a) \(\pi \int_{0}^{4}(8-2 x)^2 d x=2 \pi \int_{0}^{8}y\left(4-\frac{y}{2}\right) d y\) (b) \(\int_{0}^{2}\left(25-\left(x^2+1\right)^2\right) d x=2 \int_{1}^{5} y\sqrt{y-1} d y\) Answer: (a) The equality is true because the functions \((8-2x)^2\) and \(y(4-\frac{y}{2})\) are related through a coordinate transformation, and their integrals have the same function with transformed limits of integration. (b) The equality is true because the integrals can be split into terms that cancel out, leaving the remaining terms equivalent through a suitable substitution.

Step 1: Sketch the graphs

Sketch the graph of \((8-2x)^2\) and \(y(4-\frac{y}{2})\) in separate Cartesian coordinate systems.

Step 2: Transform the coordinate systems

Notice that the graph of \((8-2x)^2\) is a parabola in x-y plane opening downwards and the graph of \(y(4-\frac{y}{2})\) is a parabola in x-y plane opening to the right. To find a relation between these two functions, we apply a coordinate transformation. Make a change of variables: \(x \to 4 - \frac{y}{2}, \; y \to 8 - 2x\).

Step 3: Calculate the new function

Make this substitution in the expression \((8-2x)^2\): $$(8-2x)^2 = \left(8-2(4-\frac{y}{2})\right)^2 = \left(\frac{y}{2}\right)^2 = y\left(4-\frac{y}{2}\right)$$

Step 4: Relate the integrals

Now, notice that the integrals have the same function, differing only by their limits of integration. We can see that the limits of the first integral are transformed to the limits of the second integral through the coordinate transformation: $$x: 0 \to 8; \quad x: 4 \to 0$$ Therefore, the equality is true. We can conclude that: $$\pi \int_{0}^{4}(8-2 x)^2 d x=2 \pi \int_{0}^{8}y\left(4-\frac{y}{2}\right) d y$$ For part (b):

Step 1: Observe the given functions

Notice that the given functions are \(25-(x^2+1)^2\) and \(y\sqrt{y-1}\).

Step 2: Apply a suitable substitution

Let's set \(t = x^2 + 1\), which implies \(dt = 2x\,dx\). Then, the given equality can be rewritten as: $$\int_{(0^2+1)}^{(2^2+1)}\left(25-t^2\right) \frac{dt}{2\sqrt{t-1}}=2 \int_{1}^{5} y\sqrt{y-1} d y$$

Step 3: Simplify the integrals

Comparing the two integrals, we have: $$\frac{1}{2}\int_{1}^{5} \frac{25-t^2}{\sqrt{t-1}}dt = 2 \int_{1}^{5} y\sqrt{y-1} dy$$ Further simplify the left integral: $$\frac{1}{2}\int_{1}^{5} \frac{(5-t)(5+t)}{\sqrt{t-1}}dt = 2 \int_{1}^{5} y\sqrt{y-1} dy$$ Now, notice how the left side of the equation has the term \((5-t)\). In order to establish the equality, we can divide the left side into two terms: $$\frac{1}{2}\int_{1}^{5} \frac{5-t}{\sqrt{t-1}}dt + \frac{1}{2}\int_{1}^{5} \frac{5+t}{\sqrt{t-1}}dt = 2 \int_{1}^{5} y\sqrt{y-1} dy$$

Step 4: Establish the equality

The term \(\frac{5-t}{\sqrt{t-1}}\) is equivalent to \(y\sqrt{y-1}\) when \(t = y\). The term \(\frac{5+t}{\sqrt{t-1}}\) is the opposite of \(y\sqrt{y-1}\) when \(t = 5 - y\). By comparing these relations, we observe that the second term on the left side cancels out, and the equality holds true: $$\int_{0}^{2}\left(25-\left(x^2+1\right)^2\right) d x=2 \int_{1}^{5} y\sqrt{y-1} d y$$

Key Concepts

Coordinate Transformation

In mathematics, coordinate transformations are used to translate a problem from one coordinate system to another. This technique often simplifies complex problems, making them easier to solve.
For instance, transforming Cartesian coordinates to polar coordinates can make certain integrals more manageable.
In our exercise, we observe how changing variables can relate two apparently different integrals. By replacing

• \( x \) with \(4 - \frac{y}{2}\)
• \( y \) with \(8 - 2x\)

we see that two different-looking functions become identical. Thus, their integrals are also equal, just with different limits. Transforming coordinates helps reveal such underlying similarities by adjusting perspectives.

Graphical Interpretation

Understanding integrals through graphical interpretation involves visualizing the geometrical shape of the region represented by the integral.
Sketching graphs of functions like \((8-2x)^2\) and \(y(4-\frac{y}{2})\) highlights their parabolic nature.
This gives insight into the areas under these curves.
When coordinates are transformed, the function's graph also undergoes a change. By visualizing these transformations, we understand how areas and boundaries shift.
Our exercise illustrates that despite different algebraic representations, the areas encapsulated by these graphs are equivalent after transformation.

Parabolic Functions

Parabolas appear frequently in mathematics, particularly in functions such as \((8-2x)^2\) and \(y(4-\frac{y}{2})\).
A parabolic function typically takes the form \(ax^2 + bx + c\), and these are defined by their vertex, axis of symmetry, and direction of opening.
In our context, each function describes a different orientation of parabola:

• \((8-2x)^2\) represents a downward-opening parabola when sketched on an x-y plane.
• \(y(4-\frac{y}{2})\) signifies a parabola opening to the right when sketched.

These visual and algebraic characteristics help in matching and relating integrals from the exercise, especially through the process of transformation. Understanding parabolic functions aids in interpreting integral values and relationships geometrically.