Finding Area Between Curves \(y=4x^2\) and \(y=4\)

To find the area of the region enclosed between the curves \(4x^2 = y\) and \(y = 4\), we need to follow these steps:

1. Identify the equations of the curves.
2. Find the points where the curves intersect.
3. Determine which curve is above the other in the region between the intersection points.
4. Set up and evaluate the definite integral representing the area.

Identifying the Curves

• The first curve is given by \(4x^2 = y\), which can be written as \(y = 4x^2\). This is the equation of a parabola opening upwards with its vertex at the origin \((0,0)\).
• The second curve is given by \(y = 4\). This is the equation of a horizontal line passing through \(y=4\).

Finding Intersection Points

The curves intersect where their y-values are equal. So, we set the two equations equal to each other:

\(4x^2 = 4\)

Divide both sides by 4:

\(x^2 = 1\)

Taking the square root of both sides gives the x-coordinates of the intersection points:

\(x = \pm \sqrt{1}\)

\(x = \pm 1\)

The intersection points occur at \(x = -1\) and \(x = 1\).

Setting Up the Integral for Area

The region enclosed by the curves lies between \(x = -1\) and \(x = 1\). We need to determine which curve forms the upper boundary and which forms the lower boundary within this interval.

Let's test a point within the interval \((-1, 1)\), for example, \(x = 0\).

• For the line \(y = 4\), \(y = 4\).
• For the parabola \(y = 4x^2\), \(y = 4(0)^2 = 0\).

Since \(4 > 0\) at \(x=0\), the line \(y=4\) is above the parabola \(y=4x^2\) throughout the interval \([-1, 1]\).

The area \(A\) of the region between two curves \(y=f(x)\) and \(y=g(x)\) from \(x=a\) to \(x=b\), where \(f(x) \ge g(x)\) on \([a, b]\), is given by:

\(A = \int_{a}^{b} (f(x) - g(x)) dx\)

In our case, \(f(x) = 4\), \(g(x) = 4x^2\), \(a = -1\), and \(b = 1\). So the integral is:

\(A = \int_{-1}^{1} (4 - 4x^2) dx\)

Evaluating the Definite Integral

Now, we evaluate the definite integral to find the area:

First, find the antiderivative of \(4 - 4x^2\):

\(\int (4 - 4x^2) dx = 4x - \frac{4x^{2+1}}{2+1} + C = 4x - \frac{4x^3}{3} + C\)

Now, evaluate the definite integral from -1 to 1:

\(A = \left[ 4x - \frac{4x^3}{3} \right]_{-1}^{1}\)

Apply the limits of integration (Upper limit - Lower limit):

\(A = \left( 4(1) - \frac{4(1)^3}{3} \right) - \left( 4(-1) - \frac{4(-1)^3}{3} \right)\)

\(A = \left( 4 - \frac{4}{3} \right) - \left( -4 - \frac{4(-1)}{3} \right)\)

\(A = \left( 4 - \frac{4}{3} \right) - \left( -4 + \frac{4}{3} \right)\)

Now, simplify the terms:

\(A = 4 - \frac{4}{3} + 4 - \frac{4}{3}\)

\(A = (4 + 4) - \left( \frac{4}{3} + \frac{4}{3} \right)\)

\(A = 8 - \frac{8}{3}\)

Combine the terms with a common denominator:

\(A = \frac{8 \times 3}{3} - \frac{8}{3} = \frac{24}{3} - \frac{8}{3}\)

\(A = \frac{24 - 8}{3}\)

\(A = \frac{16}{3}\)

Alternatively, using the property of even functions (since \(4 - 4x^2\) is even and the interval is symmetric \([-1, 1]\)):

\(A = 2 \int_{0}^{1} (4 - 4x^2) dx\)

\(A = 2 \left[ 4x - \frac{4x^3}{3} \right]_{0}^{1}\)

\(A = 2 \left[ \left( 4(1) - \frac{4(1)^3}{3} \right) - \left( 4(0) - \frac{4(0)^3}{3} \right) \right]\)

\(A = 2 \left[ \left( 4 - \frac{4}{3} \right) - (0 - 0) \right]\)

\(A = 2 \left[ \frac{12 - 4}{3} \right]\)

\(A = 2 \left[ \frac{8}{3} \right]\)

\(A = \frac{16}{3}\)

Final Area Calculation

The calculated area of the region enclosed between the curves \(4x^2 = y\) and \(y = 4\) is \(\frac{16}{3}\) square units.

Revision Table: Area Calculation Steps

Additional Information: Calculus Area Concepts

Finding the area between curves is a fundamental application of definite integrals in calculus. Here are some related concepts:

• Definite Integral: Represents the signed area between a curve and the x-axis over a given interval. The integral \(\int_{a}^{b} f(x) dx\) calculates this area.
• Area Between Curves: If \(f(x)\) and \(g(x)\) are continuous functions on \([a, b]\) and \(f(x) \ge g(x)\) for all \(x\) in \([a, b]\), the area between the curves from \(x=a\) to \(x=b\) is \(\int_{a}^{b} (f(x) - g(x)) dx\).
• Integration with Respect to y: Sometimes, it is easier to find the area by integrating with respect to y. If the region is bounded by curves \(x=f(y)\) and \(x=g(y)\) from \(y=c\) to \(y=d\), the area is \(\int_{c}^{d} |f(y) - g(y)| dy\).
• Symmetry: For regions symmetric about the y-axis (like the one in this problem) and integrals over symmetric intervals \([-a, a]\), the property \(\int_{-a}^{a} f(x) dx = 2 \int_{0}^{a} f(x) dx\) can simplify calculations if \(f(x)\) is an even function (\(f(-x) = f(x)\)).