Problem 29 - Calculus Early Transcendentals

Question

Net area and definite integrals Use geometry (not Riemann sums) to evaluate the following definite integrals. Sketch a graph of the integrand, show the region in question, and interpret your result. $$\int_{0}^{4} \sqrt{16-x^{2}} d x$$

Step-by-Step Solution

Answer

The area under the curve of the function $$\sqrt{16-x^2}$$ in the interval $$[0, 4]$$ represents the net area of the region enclosed between the curve, x-axis, and the bounds $$0$$ and $$4$$. In this particular case, the region forms a quarter-circle shape.

Step 1: Sketch the graph

First, let's sketch the graph of the function $$\sqrt{16-x^{2}}$$ in the given interval $$[0, 4]$$. This function represents the upper half of a circle with radius 4 and centered at the origin. In our interval, we will see that the graph is a quarter of a circle.

Step 2: Find the geometric shape

To find the net area under the curve, we need to inspect the shape formed by the region enclosed between the function, x-axis, and the bounds $$0$$ and $$4$$. The region forms a quarter-circle shape, and hence we can calculate its area by finding the area of the whole circle and divide it by 4.

Step 3: Calculate the area

To calculate the area of the whole circle, we use the formula: $$A = \pi r^{2}$$, where $$A$$ is the area, $$\pi$$ is a constant, and $$r$$ is the radius (in our case, r = 4). So, we have $$A = \pi (4)^{2} = \pi*16$$ Since the net area we are trying to find is the area of the quarter-circle, we divide the total area by $$4$$. $$A_{quartercircle}=\frac{\pi*16}{4}$$ $$A_{quartercircle}=4\pi$$

Step 4: Interpret the result

The area under the curve of the function $$\sqrt{16-x^{2}}$$ in the interval $$[0, 4]$$, equals $$4\pi$$. This means that the region enclosed between the function, x-axis, and the bounds $$0$$ and $$4$$ has a net area of $$4\pi$$.

Key Concepts

Net Area

The net area in the context of definite integrals represents the total area between the graph of a function and the x-axis, over a given interval. It is important to note that if the graph falls below the x-axis, the area is considered negative, which might affect the net area. But in our specific exercise, the function \( \sqrt{16-x^{2}} \) remains above the x-axis for the interval \([0, 4]\). Therefore, the net area is simply the area under the curve.

To compute this, instead of using cumbersome Riemann sums, which are sums of infinite small rectangles to approximate the area, we use a geometric approach. This is much more straightforward when dealing with shapes like circles or triangles, which have well-known formulas. For this problem, we rely on recognizing that the shape is part of a circle and use the corresponding formula to find the net area efficiently.

Geometric Interpretation

Geometric interpretation provides an intuitive understanding of definite integrals. It helps visualize what the math is looking to compute. In this exercise, the function \( \sqrt{16-x^{2}} \) forms the upper half of a circle with a radius of 4. This is a direct result of the equation \( x^2 + y^2 = r^2 \), where this function is equivalent to rearranging to find y.

Specifically, the portion of the circle that lies in the interval \([0, 4]\)on the x-axis is a quarter of the entire circle. By interpreting this as a geometrical shape — a quarter-circle — we can simplify the solution process. The understanding is not just algebraic but also lends a clear visual of slicing a pie into four equal parts, and focusing only on one of those slices.

• We find the total area of the circle first, using \( A = \pi r^{2} \).
• Then divide by four, since we only want the quarter region.

This approach makes it easier to solve the problem without delving deep into complex calculus calculations.

Quarter-Circle Area

The quarter-circle area in this exercise arises naturally once we identify the graph of the function \( \sqrt{16-x^{2}} \). This part of the circle is crucial, as it simplifies the computation process immensely using geometric techniques. Circles have symmetric, well-known properties, allowing their area components to be calculated easily.

Since the full circle with radius 4 has an area of \( \pi (4)^2 = 16\pi \), dividing by 4 gives the area of one quarter of the circle:

• \( A_{quartercircle} = \frac{\pi \times 16}{4} \)
• \( A_{quartercircle} = 4\pi \)

This \( 4\pi \)is what we find as the definite integral value over the interval \([0, 4]\). Understanding this, we see that it's not just an algebraic manipulation; it's about identifying the nature of the shape and leveraging that understanding to solve the problem swiftly.