Problem 82 - Calculus Volume 1

Question

In the following exercises, evaluate the integral using area formulas. \(\int_{0}^{12} \sqrt{36-(x-6)^{2}} d x\)

Step-by-Step Solution

Answer

The integral evaluates to \(18\pi\).

Step 1: Identify the Function

The function inside the integral is \ \( y = \sqrt{36-(x-6)^2} \). This represents a semicircle with the equation \( (x-6)^2 + y^2 = 36 \).

Step 2: Determine Semicircle Properties

The equation above can be rewritten as \( x^2 + y^2 = 36 \) by adjusting for the center. This describes a circle with radius 6, centered at \((6, 0)\). Since we have \( y = \sqrt{36-(x-6)^2} \), this represents the top half or semicircle of the circle.

Step 3: Determine the Bounds of Integration

We have the bounds \( x = 0 \) to \( x = 12 \). This range completely covers the semicircle as it spans from \( x = 0 \) to \( x = 12 \), the full diameter along the x-axis.

Step 4: Compute the Area of the Semicircle

The area of a full circle is given by \( \pi r^2 \), where \( r = 6 \). The area of this circle is \( \pi(6)^2 = 36\pi \). The semicircle, being half of the circle, has an area of \( \frac{36\pi}{2} = 18\pi \).

Step 5: Evaluate the Integral

Since the integral represents the area under the curve from \( x = 0 \) to \( x = 12 \), which matches the semicircle's area, the value of the integral is simply \( 18\pi \).

Key Concepts

Geometry in Calculus

When studying calculus, one often encounters the use of geometric shapes to understand integrals better. Geometry provides a visual insight into problems, making it easier to compute areas and solve integrals. Understanding geometry is key since it helps in recognizing the shapes defined by functions.
For instance, in the context of definite integrals, identifying a function as a part of a geometric shape, such as a semicircle, allows us to apply known area formulas. This is much simpler than trying to integrate the function manually, especially if the shape is regular like a circle or rectangle.
In the problem at hand, we analyze the function within the integral to see that it forms a semicircle. This insight directly links geometry with calculus, providing a tangible method to find solutions. Using geometric shapes to simplify interpretation and calculation in calculus provides a helpful bridge between algebraic expressions and spatial understanding.

Area under a Curve

The concept of finding the area under a curve is central to integrating functions in calculus. When we compute the definite integral of a function, we're essentially finding the area enclosed by the curve, the x-axis, and the vertical lines at the bounds of integration.

• The area under a curve can represent various practical concepts, such as distance, probability, or accumulated quantities.
• Definite integrals give a precise numerical area, providing crucial insights into the behavior of functions within a bounded interval.

When dealing with curves that describe geometric shapes, this creates opportunities to apply simple area formulas.In the given exercise, the integral of \(\int_{0}^{12} \sqrt{36-(x-6)^{2}} dx\) tells us the area under the semicircle from the x-axis between the bounds.Using known formulas for areas of circles, we find this area to be \(18\pi\).
Leveraging area formulas simplifies the process to a straightforward computation rather than a complex algebraic integration.

Semicircle Properties

A semicircle is essentially half of a circle, defined by its radius and its center on a coordinate plane. Typically, the equation \(x^2 + y^2 = r^2\) defines a full circle, while rearranging and solving for \(y\) gives one half of its boundary.

• The semicircle seen above the x-axis is represented by \(y = \sqrt{r^2 - x^2}\).
• Its area is half of the area of the circle, implying the formula is \(\frac{1}{2} \pi r^2\).

In the problem given, the function \(y = \sqrt{36-(x-6)^2}\) represents the top half of the circle, shifted to have its center at (6, 0). The radius here is 6.
The bounds for integration (0 to 12) fully capture this semicircle because its diameter stretches between \(x = 0\) and \(x = 12\).By understanding these properties, calculating the area becomes a straightforward task, reinforcing the beauty and utility of combining geometry with calculus to solve integrals.