Question

In Exercises \(15-34,\) find the area of the regions enclosed by the lines and curves. $$x-y^{2}=0 \quad$$ and $$\quad x+2 y^{2}=3$$

Short answer

The area enclosed by the curves \(y = \sqrt{x}\) and \(y = \sqrt{(3 - x) / 2}\) is found to be \(\frac{7}{18}\).

Step-by-step solution

Rewrite the equations

First, rewrite the given equations in y=f(x) format. \n\n \(x = y^2\) can be rewritten as \(y = \sqrt{x}\) \n and \(x + 2y^2 = 3\) as \(y = \sqrt{(3 - x) / 2}\)

Find points of intersection

Find the points where the two curves intersect. This means finding the x-values for which both equations are equal. \n Set \(\sqrt{x} = \sqrt{(3 - x) / 2}\), solve this equation to get the intersection points. Square both sides to get rid of the square root then solving gives x = 1,2. Meaning the regions intersect at points (1,1) and (2,1.414)

Integrate the functions

The area A between the curves is the absolute difference of the integrals between the boundaries, which are the intersection points. \n So, evaluate the integral A = \(\int_{1}^{2} (\sqrt{x} - \sqrt{(3-x)/2}) dx\). Break the integral into two separate integrals and solve each one. The antiderivative of \(\sqrt{x}\) is \(\frac{2}{3}x^\frac{3}{2}\) and the antiderivative of \(\sqrt{(3-x)/2}\) is -\(\frac{2}{3}(3-x)^\frac{3}{2}\). Use the Fundamental Theorem of Calculus to evaluate the definite integrals from 1 to 2.

Compute the area

Substitute the limits into the antiderivative determined in the previous step and compute the difference. This will give the area enclosed by the curves.

Key concepts

Integration Techniques

When we talk about integration techniques in calculus, we're referring to the various methods used to evaluate integrals. Since integration is a key part of finding the area between curves, mastering these techniques is crucial for solving such problems. For instance, in the exercise provided, we have the task of integrating functions like \( \sqrt{x} \) and \( \sqrt{(3-x)/2} \).

To deal with these square roots, one commonly used integration technique is substitution, which simplifies the function into a more manageable form. However, in this case, we do not need substitution as the power function rule can be applied directly. This rule states that when integrating a function of the form \( x^n \), where \( n \) is not equal to \( -1 \), the integral is \( \frac{x^{n+1}}{n+1} \), plus the constant of integration.

Applying this rule to \( \sqrt{x} = x^{\frac{1}{2}} \) gives us the antiderivative \( \frac{2}{3}x^{\frac{3}{2}}\). On the other hand, the presence of a binomial \( (3-x) \) under a square root in \( \sqrt{(3-x)/2} \) indicates that more attention must be paid to signs during integration. The antiderivative found here is \( -\frac{2}{3}(3-x)^{\frac{3}{2}} \). Balancing these antiderivatives and understanding when to use these techniques forms the foundation of the integration process.

Definite Integrals

Definite integrals are not just about computing the accumulation of quantities but play a fundamental role in finding the exact area under a curve between two points. We use definite integrals when we want to calculate the area under a curve from one point, say \( a \), to another, \( b \). Our example problem requires us to calculate the area between two curves making use of definite integrals.

In practical terms, to find the area between the curves \( y = \sqrt{x} \) and \( y = \sqrt{(3 - x) / 2} \) from \( x=1 \) to \( x=2 \) we set up the definite integral \( \int_{1}^{2} (\sqrt{x} - \sqrt{(3-x)/2}) dx\). This means we're looking for the total 'signed' area between the two curves over this interval. 'Signed' here refers to the fact that we consider the relative position of the curves: the area is positive if the upper curve lies above the lower curve and negative if it is below.

The computation of a definite integral results in a numerical value, which represents this precise area. As the area cannot be negative, we often take the absolute difference of the integrals to ensure our solution reflects physical reality—positive areas.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus is a stunning piece of mathematical work that bridges the concepts of differentiation and integration, essentially showing that they are inverse processes. This theorem comes in two parts: the first part establishes that if a function \( f \) is continuous over an interval and \( F \) is its antiderivative, then \( F \) can be used to evaluate the definite integral of \( f \) over that interval. In the context of our area problem, after finding the antiderivatives of \( \sqrt{x} \) and \( \sqrt{(3 - x) / 2} \), the theorem indicates how to calculate the definite integrals that represent the areas under each curve.

The second part of the theorem deals with the evaluation. It states that if you want to evaluate \( \int_{a}^{b} f(x) dx \), you can simply compute \( F(b) - F(a) \) where \( F \) is the antiderivative of \( f \). This subtraction gives us the net area between the curves from \( a \) to \( b \). For the exercise, applying this part of the theorem to our antiderivatives \( \frac{2}{3}x^{\frac{3}{2}} \) and \( -\frac{2}{3}(3-x)^{\frac{3}{2}} \) and evaluating them from \( x=1 \) to \( x=2 \) provides us with the exact area between the given curves. In simple terms, the Fundamental Theorem of Calculus gives us the power to perform practical calculations of areas with confidence in the results.