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Chapter 11: Problem 94

Find the area of the region enclosed by the graphs of \(y=\sqrt{9-x^{2}}\) and \(y=x+3\)

Short Answer

Expert verified
Enclosed area={9}+6

Step by step solution

01

Identify Intersection Points

The intersection points are found by setting the two equations equal: y=yeqy=sqrt{9-x^{2}}x^3-(x+3). Solve the equation sqrt{9-x^{2}}=x+3x^3-(x+3). Squaring both sides and simplifying gives the quadratic equation x^{2}+6x-0=0 leading to x^{2}+6x-0=(x-3)(x+3). Solving this equation provides the intersection points.
02

Evaluate Definite Integral

Determine the integral of the difference of the two functions over the interval [3,3]x^{3}(sqrt{9-x^{2}}-(x+3))dx+=0eq. Integrate to find the area enclosed by y=sqrt{9-x^{2}}=(x+3).

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Key Concepts

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

Definite Integral
A definite integral helps calculate the area under a curve between two points on the x-axis. When dealing with functions, it is particularly useful in finding the area between curves. For our problem, we are interested in the area between the curves described by the functions: \( y = \sqrt{9 - x^2} \) and \( y = x + 3 \).

  • First, identify the points of intersection since they define the range for integration.
  • Then, subtract the lower function from the upper function within the interval of these points.
The definite integral is written as: \[ \text{Area} = \int_{a}^{b} [f(x) - g(x)] \, dx \] Here, \( f(x) = \sqrt{9 - x^2} \) and \( g(x) = x + 3 \). Integration is needed over the interval between the intersection points.
Intersection Points
Intersection points help define the limits for the definite integral. They are the x-values where the two curves intersect.

To find these points, set the equations equal to each other: \[ \sqrt{9 - x^2} = x + 3 \] Square both sides to eliminate the square root: \[ 9 - x^2 = (x + 3)^2 \] Simplify and solve the resulting quadratic equation: \[ 9 - x^2 = x^2 + 6x + 9 \] \[ 0 = 2x^2 + 6x \] This simplifies to: \[ x(x + 3) = 0 \] Hence, the intersection points are: \( x = 0 \) and \( x = -3 \). These define the integral limits for our area calculation.
Quadratic Equation
A quadratic equation is a second-degree polynomial equation of the form \( ax^2 + bx + c = 0 \). It is solved using the quadratic formula or by factoring.

In our problem, the quadratic equation arose from setting \( \sqrt{9 - x^2} = x + 3 \) and then squaring both sides, which yielded: \[ 9 - x^2 = x^2 + 6x + 9 \] This simplifies to: \[ 0 = 2x^2 + 6x \]
  • Factoring out the common term x:
    • \[ x(2x + 6) = 0 \]
Solving for x gives the roots: \( x = 0 \) and \( x = -3 \).
Solved quadratic equations determine where the curves intersect on the x-axis, which is crucial for defining the interval for our definite integral.

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