Question

Sketch the region bounded by the graphs of the given equations, show a typical slice, approximate its area, set up an integral, and calculate the area of the region. Make an estimate of the area to confirm your answer. \(x=y^{2}-2 y, x-y-4=0\)

Short answer

The area of the region is \( \frac{133}{3} \) square units.

Step-by-step solution

Find Points of Intersection

To find where the curves intersect, set the equations equal: \[ y^2 - 2y = y + 4. \]Rearrange to form a quadratic equation:\[ y^2 - 3y - 4 = 0. \]Factoring gives:\[ (y-4)(y+1) = 0. \]Thus, the points of intersection are at \( y = 4 \) and \( y = -1 \). Plugging these back into one of the original equations, we find the x-values at these points. At \( y = 4 \), \( x = 8 \), and at \( y = -1 \), \( x = -5 \). Therefore, the points of intersection are \((8, 4)\) and \((-5, -1)\).

Sketch the Region and Typical Slice

The equation \( x = y^2 - 2y \) describes a parabola opening to the right, and \( x = y + 4 \) is a straight line. Roughly sketch the parametric curves and note their intersection points. The region bounded is primarily between \( y = -1 \) and \( y = 4 \) along the y-axis.To approximate the area, consider a horizontal slice of thickness \( dy \). The length of this small slice is from the parabola to the line, given by:\[ \text{Length} = (y + 4) - (y^2 - 2y) = 2y - y^2 + 4. \]

Set up the Integral for Area

With the typical slice calculated, set up the integral that represents the area of the region:\[ \int_{-1}^{4} \left[(y + 4) - (y^2 - 2y)\right] \, dy = \int_{-1}^{4} (2y - y^2 + 4) \, dy. \]

Calculate the Area via Integration

Integrate the expression:\[ \int (2y - y^2 + 4) \, dy = \left[ y^2 - \frac{y^3}{3} + 4y \right]_{-1}^{4}. \]Evaluate the definite integral by substituting the bounds:\[ \left[ (16 - \frac{64}{3} + 16) - (1 + \frac{1}{3} - 4) \right]. \]Calculate, which simplifies to:\[ \frac{125}{3} - \frac{-8}{3} = \frac{133}{3}. \]Thus, the area of the region is \( \frac{133}{3} \) square units.

Estimate the Area for Confirmation

Visually estimate the area using a sketch. Approximate the average height of the region and width of slices over \([-1, 4]\). Calculation yields about 44, which should agree closely to \( \frac{133}{3} \) (approximately 44.33) from the integral, confirming the calculation is likely correct.

Key concepts

Intersection Points

To find where two curves intersect, we need to solve for the values where both equations are equal. This step is crucial because it tells us the boundaries of the area we're interested in calculating. In our problem, the two equations are:

• \(x = y^2 - 2y\)
• \(x = y + 4\)

By setting these equal to each other, we form a quadratic equation:\(y^2 - 2y = y + 4\). Rearranging gives \(y^2 - 3y - 4 = 0\), which we solve by factoring as \((y-4)(y+1)=0\). This gives the solutions \(y = 4\) and \(y = -1\). For these \(y\)-values, we find corresponding \(x\)-values which are \(x = 8\) and \(x = -5\) respectively. Therefore, the points of intersection are \((8, 4)\) and \((-5, -1)\). These points define the limits of integration for our area calculation.

Integral Setup

After finding the points of intersection, the next step is to set up the integral that will allow us to calculate the area between the curves. First, we need to visualize the setup by sketching the region and determining how the curves intersect visually.The equation \(x = y^2 - 2y\) is a parabola that opens to the right, and \(x = y + 4\) is a straight line. The region we're interested in is between these curves from \(y = -1\) to \(y = 4\).To calculate the area, we use a typical horizontal slice of thickness \(dy\). The length of this slice is the horizontal distance from the parabola to the line, given by:\[\text{Length} = (y + 4) - (y^2 - 2y) = 2y - y^2 + 4.\]This expression forms the integrand for our area calculation.

Definite Integral

The process of integration involves setting up the integral with calculated boundaries and solving it to find the area. To find the area between the curves, we integrate the expression from \(y = -1\) to \(y = 4\):\[\int_{-1}^{4} (2y - y^2 + 4) \, dy.\]Performing the integration, we find the antiderivative:\[ y^2 - \frac{y^3}{3} + 4y. \]We then evaluate this antiderivative at the upper and lower bounds of the integration:

• Upper bound at \(y = 4\): \((16 - \frac{64}{3} + 16)\)
• Lower bound at \(y = -1\): \((1 + \frac{1}{3} - 4)\)

The area is then the difference of these, which simplifies to \(\frac{133}{3}\) square units.

Region Bounded by Curves

Understanding the region bounded by curves comes from identifying the common area enclosed by these curves. In this example, the curves \(x = y^2 - 2y\) and \(x = y + 4\) enclose a particular segment in the plane.The region is traversed vertically from the intersections at \(y = -1\) to \(y = 4\). The area is defined as the portion where one curve is consistently above the other within these bounds.Visualizing this helps in confirming the integration setup. Pencil out sketches can support seeing precisely how these curves intersect, ensuring the calculations reflect the actual enclosed region.In summary, fully grasping the region bounded by curves is crucial for correctly applying integral calculus to determine the area between them.