Problem 35 - Calculus Single Variable

Question

Treat the \(y\) variable as the independent variable and the \(x\) variable as the dependent variable. By integrating with respect to \(y,\) calculate the area of the region that is described. The region between the curves \(x=y^{2}\) and \(x=-y^{2}+4\)

Step-by-Step Solution

Answer

The area between the curves is \(\frac{16\sqrt{2}}{3}\).

Step 1: Identify the points of intersection

To find the region between the curves, we need to determine where the two curves intersect. Set the equations equal to each other: \(x = y^2\) and \(x = -y^2 + 4\). Equal these two expressions: \(y^2 = -y^2 + 4\). Solve for \(y\): \[2y^2 = 4\] Simplifying gives \(y^2 = 2\), thus \(y = \sqrt{2}\) or \(y = -\sqrt{2}\). The curves intersect at \(y = -\sqrt{2}\) and \(y = \sqrt{2}\).

Step 2: Set up the integral for the area

The area between the curves is found by integrating the difference between the outer curve and the inner curve with respect to \(y\) from \(-\sqrt{2}\) to \(\sqrt{2}\). Here, the curve \(x = -y^2 + 4\) is the outer curve and \(x = y^2\) is the inner curve. Thus, the integral becomes: \[\int_{-\sqrt{2}}^{\sqrt{2}} ( (-y^2 + 4) - y^2 ) \, dy\] Simplifying the expression inside the integral gives: \[\int_{-\sqrt{2}}^{\sqrt{2}} (-2y^2 + 4) \, dy\]

Step 3: Integrate with respect to y

Evaluate the integral: \[\int (-2y^2 + 4) \, dy\] This equals \(-\frac{2}{3}y^3 + 4y\). Now compute this from \(-\sqrt{2}\) to \(\sqrt{2}\).

Step 4: Calculate the definite integral

Substitute \(\sqrt{2}\) and \(-\sqrt{2}\) into the antiderivative: \[\left[ -\frac{2}{3}y^3 + 4y \right]_{-\sqrt{2}}^{\sqrt{2}}\]Calculate each part: \[= \left(-\frac{2}{3} (\sqrt{2})^3 + 4 \cdot \sqrt{2}\right) - \left(-\frac{2}{3} (-\sqrt{2})^3 + 4 \cdot (-\sqrt{2})\right)\]Simplify: \[= \left(-\frac{2}{3}(2\sqrt{2}) + 4\sqrt{2}\right) - \left(\frac{2}{3}(2\sqrt{2}) - 4\sqrt{2}\right)\] Solving this gives: \[= \frac{16\sqrt{2}}{3}\]

Step 5: Write the final expression for the area

The area between the curves is \(\frac{16\sqrt{2}}{3}\). This area represents the space enclosed between the two parabola-shaped curves.

Key Concepts

Definite Integral

A definite integral represents the accumulation of quantities, like area, over an interval. In this scenario, the definite integral helps calculate the area between the two curves. When you integrate from a lower to an upper bound, in this case from \(-\sqrt{2}\) to \(\sqrt{2}\), you effectively find the total 'net area' between the given limits. Here’s how it’s done:

• Identify the difference between the outer and inner curves.
• Set limits according to the points of intersection.
• Integrate the difference, which sums up the area between the curves over the interval.

This area is represented by the definite integral \(\int_{-\sqrt{2}}^{\sqrt{2}}(-2y^2+4)\,dy\). The result gives you the total area trapped between the two curves over this specified interval.

Area Between Curves

Finding the area between two curves involves determining which function lies above the other in the region of interest. This exercise simplifies this by treating \(y\) as the variable of integration.
The basic steps include:

• Solving for the points of intersection to decide the limits of integration.
• Determining which function is the outermost curve along those limits.
• Subtracting the inner from the outer curve yields an expression which can be integrated to find the area:
  \((-y^2 + 4) - y^2 = -2y^2 + 4\).

This expression is integrated to find the area between \(x = y^2\) and \(x = -y^2 + 4\), which are two parabolic curves, thus creating an enclosed region whose area is being calculated.

Points of Intersection

Points of intersection are crucial when dealing with areas between curves as they define the boundaries of the region where areas need to be calculated.
Here's how you find them:

• Set the equations equal: \(y^2 = -y^2 + 4\).
• Solve for \(y\). This led to \(2y^2 = 4\) yielding \(y = \sqrt{2}\) or \(y = -\sqrt{2}\).

These solutions tell you precisely where the two curves meet. This point of intersect is important because it indicates where the area calculation should start and stop. Ensuring accuracy in these calculations is key to finding the right area.

Indefinite Integral

An indefinite integral, unlike its definite counterpart, does not have bounds and represents a family of functions. Here, we deal with an indefinite integral to find the antiderivative of the given expressions:
\[\int (-2y^2 + 4) \, dy\].

• This gives an antiderivative: \(-\frac{2}{3}y^3 + 4y\).
• This antiderivative is then evaluated at the limits \(-\sqrt{2}\) and \(\sqrt{2}\) to determine the definite integral.

Thus, the indefinite integral provides a means of expressing the accumulated value, in terms of an antiderivative, necessary for the computation of the specific area between the given curves. This highlights its foundational role in solving definite integrals for practical applications.