Problem 76 - Thomas Calculus

Question

Find the area of the region between the curve \(y=3-x^{2}\) and the line \(y=-1\) by integrating with respect to a. \(x, \quad\) b. \(y\) .

Step-by-Step Solution

Answer

The area is \(\frac{32}{3}\).

Step 1: Identify Intersection Points (x-axis integration)

To find the area between the curve \(y = 3 - x^2\) and the line \(y = -1\), we need to identify the points where these two equations intersect. Equate \(3 - x^2 = -1\). Solving, we get \(3 = x^2 - 1\) or \(x^2 = 4\). Thus, \(x = \pm 2\). This means the intersection points are \((-2, -1)\) and \((2, -1)\).

Step 2: Set Up the Integral (x-axis integration)

We integrate the difference of the functions from \(x = -2\) to \(x = 2\). The area \(A\) between the curve \(3-x^2\) and the line \(-1\) is given by the integral: \[A = \int_{-2}^{2} ((3 - x^2) - (-1))\, dx = \int_{-2}^{2} (4 - x^2)\, dx\]

Step 3: Compute the Integral (x-axis integration)

Integrate \( \int (4 - x^2)\, dx \): \[\int (4 - x^2)\, dx = 4x - \frac{x^3}{3}\]from \(-2\) to \(2\). Substitute:\[\left[4x - \frac{x^3}{3}\right]_{-2}^{2} = \left(4(2) - \frac{(2)^3}{3}\right) - \left(4(-2) - \frac{(-2)^3}{3}\right)\]This simplifies to:\[\left(8 - \frac{8}{3}\right) + \left(8 + \frac{8}{3}\right) = 16 - \frac{16}{3} = \frac{48}{3} - \frac{16}{3} = \frac{32}{3}\]

Step 4: Identify Intersection Points (y-axis integration)

The curve \(y = 3 - x^2\) implies that \(x^2 = 3 - y\), and thus \(x = \pm \sqrt{3-y}\). The intersection points when \(y = -1\) are already \(-2\) and \(2\) in terms of \(x\), which means valid ranges for \(y\) are \([-1, 3]\).

Step 5: Set Up the Integral (y-axis integration)

We need to calculate the area between left and right bounds of \(x\) values that come from \(x = -\sqrt{3-y}\) and \(x = \sqrt{3-y}\). This area is\[A = \int_{-1}^{3} (\sqrt{3-y} - (-\sqrt{3-y}))\, dy = 2\int_{-1}^{3} \sqrt{3-y} \ dy\]

Step 6: Compute the Integral (y-axis integration)

Integrate: \[2\int \sqrt{3-y}\, dy\]which simplifies to\[2\left[-\frac{2}{3}(3-y)^{3/2}\right]_{-1}^{3}\]Evaluate from \(-1\) to \(3\):\[2\left[0 - (-\frac{2}{3}(3-(-1))^{3/2})\right]\]This evaluates to:\[2\left[\frac{2}{3}(4)^{3/2}\right] = 2\left[\frac{16}{3}\right] = \frac{32}{3}\]

Step 7: Conclusion: Verification and Area

Both methods yield the same area \(\frac{32}{3}\), confirming that the area of the region between the curve \(y = 3 - x^2\) and the line \(y = -1\) is indeed \(\frac{32}{3}\).

Key Concepts

Definite Integrals

A definite integral is a powerful tool to find the exact area between curves on a specified interval. Imagine you have two functions, like in our exercise: the curve \(y = 3 - x^2\) and the line \(y = -1\). By integrating the area between them over a specific range of \(x\) or \(y\), you can determine the precise size of the region they encapsulate.

This process involves setting up the integral based on the intersections you find. For instance, when integrating with respect to \(x\), the integral would be from \(x = -2\) to \(x = 2\). The definite integral will compute the entire area between these two \(x\) limits. The formula becomes \(\int_{-2}^{2} ((3 - x^2) - (-1)) \, dx = \int_{-2}^{2} (4 - x^2) \, dx\).

When integrating with respect to \(y\), the resulting integral changes to \( \int_{-1}^{3} 2 \sqrt{3-y} \, dy \). Integration is performed over the interval to sum up tiny slices of area which are calculated using simple functions. The definite integral then provides us with the total area, translated from our mathematical model into a real-world quantity: \( \frac{32}{3} \). This result is effectively visualized as the space between the curve and the line.

Coordinate Geometry

Coordinate geometry helps us handle complex problems like finding areas between curves by using a clear set of tools and definitions. Consider coordinates as our way to navigate mathematical space. For example, the curve \(y = 3 - x^2\) and the line \(y = -1\) are plotted within a coordinate system, allowing us to track their paths and interactions.

Based on their equations, coordinate geometry lets us pinpoint where they meet. These intersection points provide the bounds for our integration.

• For integration with respect to \(x\), we use the points \((-2, -1)\) and \((2, -1)\).
• For integration with respect to \(y\), the curve translates to \(x = \pm \sqrt{3-y}\).

In coordinate geometry, plotting this ensures we can visualize their overlap, crucial for determining which range to integrate over.

The coordinates help us map these functions into easier parts to evaluate, either on a plane with \(x\) or \(y\) as the focus. By using the right formulas and knowing where the functions interact — as given by the coordinates — solving for the area becomes tangible and direct.

Intersection Points

Intersection points are vital in solving problems about areas between curves. They act as the stitch between the curves, where their values match.

• To discover these points, we set the equations equal: \(3 - x^2 = -1\).
• Simplifying gives \(x^2 = 4\), resulting in \(x = \pm 2\).
• These \(x\) values tell us where the curves intersect, namely at \((-2, -1)\) and \((2, -1)\).

For integration in the \(y\)-direction, we also rearrange the curve \(y = 3 - x^2\) to get \(x = \pm \sqrt{3-y}\). Points like these set our bounds in the y-direction, from \(y = -1\) to \(y = 3\).

These intersection points form the foundation for the limits of our definite integrals. They slice the entire problem into a more digestible range, enabling us to calculate the area properly. Understanding how to find and interpret intersection points is crucial, as they ensure our area calculations reflect the true segment of overlap between the curves.