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Chapter 5: Problem 18

In Exercises \(15-22,\) graph the integrands and use known area formulas to evaluate the integrals. $$ \int_{-4}^{0} \sqrt{16-x^{2}} d x $$

Short Answer

Expert verified
The integral evaluates to \(4\pi\).

Step by step solution

01

Understand the Integrand

The integrand is given as \( \sqrt{16 - x^2} \). This is the equation of a semicircle with radius 4, centered at the origin, but only the upper half because of the square root.
02

Identify the bounds

The integration bounds are from \(-4\) to \(0\). This means we want the area under the curve from \(x = -4\) to \(x = 0\), which is the left half of the semicircle.
03

Visualize the Semicircle

Graphically, this is the top half of a circle with radius 4, centered at (0,0). The circle is defined by \(x^2 + y^2 = 16\). The portion we are interested in is the upper semicircle from \(x = -4\) to \(x = 0\).
04

Calculate the Area

The area of a full circle is \( \pi r^2 \). Here \(r = 4\), so the full circle area is \(16\pi\). Since we have a semicircle, the area is half of this, \( 8\pi \).
05

Integrate the Specific Portion

The region from \(x = -4\) to \(x = 0\) is exactly half of the semicircle (since semicircle is divided into two equal halves by y-axis), so the area of this region is also half of the semicircle. Therefore, the desired area is \([0.5 \times 8 \pi] = 4\pi \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Semicircle
A semicircle is half of a circle. When you think about a circle, imagine cutting it through the middle, linearly across the diameter. What remains is a semicircle. Commonly, we consider the upper or lower halves. In our specific example, the equation \( \sqrt{16 - x^2} \) represents the top half of a circle. Visually, when we plot this function, it appears as the upper arc of a circle on the coordinate plane.
  • The semicircle has a radius, \(r = 4\).
  • The center of the circle is at the origin, \((0,0)\).
  • This forms a part of the full circle equation: \(x^2 + y^2 = r^2\).
For our semicircle, the circle equation becomes \(x^2 + y^2 = 16\). Using the square root ensures we're only considering the upper half because square roots yield only positive values.
Area Under a Curve
The concept of "area under a curve" is a key part of integral calculus. It refers to the space under a specific curve, along the x-axis. In practical terms, it's often about finding areas of unusual shapes, which aren't simple rectangles or triangles.In integral calculus, this 'area' is calculated between two specific intervals on the x-axis. You'll find this area when integrating a curve. In this exercise, we're examining the semicircle curve from \(x = -4\) to \(x = 0\).
  • The curve is defined by the function \( \sqrt{16 - x^2} \).
  • The area under this part of the curve is the area of interest.
To calculate, visualize this section as the left half of a semicircle. Since we are dealing with a symmetrical shape, finding the area becomes significantly easier using known geometrical formulas.
Definite Integral
The definite integral is a fundamental concept in calculus, used to calculate the area under a curve between two points. It essentially unifies the process of slicing up an area into infinitesimally small parts and summing these to find the total area.For a definite integral:
  • You have specific boundaries. Here, it's from \(x = -4\) to \(x = 0\).
  • These boundaries tell us the region over which we are finding our total area.
  • The symbol \( \int_{a}^{b} f(x)\,dx \) is used, where \(a\) and \(b\) are the boundaries, and \(f(x)\) is the function of the curve.
In our exercise, \( \int_{-4}^{0} \sqrt{16 - x^2}\, dx \), represents finding the area under the specific semicircle segment. Since this is a well-defined simple shape (a semicircle), traditional geometry provides a quick solution. The semicircle's full area is \(8\pi\), and half of that, representing our defined bounds, comes to \(4\pi\). This showcases how integral calculus elegantly applies to geometry.

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