Problem 9 - Thomas Calculus

Question

The integrals and sums of integrals in Exercises 9–14 give the areas of regions in the xy-plane. Sketch each region, label each bounding curve with its equation, and give the coordinates of the points where the curves intersect. Then find the area of the region. $$ \int_{0}^{6} \int_{y^{2} / 3}^{2 y} d x d y $$

Step-by-Step Solution

Answer

The area of the region is 12 square units.

Step 1: Understand the Integral

The given double integral \( \int_{0}^{6} \int_{y^{2} / 3}^{2 y} d x d y \) represents the area of a region in the xy-plane. The integral is bounded vertically from \( y = 0 \) to \( y = 6 \). For each fixed \( y \), \( x \) varies from \( \frac{y^{2}}{3} \) to \( 2y \).

Step 2: Find Intersection Points

To find the intersection points of the curves \( x = \frac{y^{2}}{3} \) and \( x = 2y \), set them equal: \[ \frac{y^{2}}{3} = 2y \] Simplify and solve the equation:\[ y^{2} = 6y \]\[ y(y - 6) = 0 \] The solutions are \( y = 0 \) and \( y = 6 \). Using these \( y \)-values in either equation gives intersection points, \((0, 0)\) and \((12, 6)\).

Step 3: Sketch the Region

Draw the curves \( x = \frac{y^{2}}{3} \) and \( x = 2y \) on an xy-plane. The parabola opens to the right, and the line goes through the origin with a slope of 2. The region of integration is bounded by these curves from \( y = 0 \) to \( y = 6 \). Highlight the region of interest between \( y = 0 \) and \( y = 6 \) where \( \frac{y^{2}}{3} \leq x \leq 2y \).

Step 4: Calculate the Area

Evaluate the integral:\[ \int_{0}^{6} \int_{y^{2} / 3}^{2 y} d x \, d y = \int_{0}^{6} \left[ x \right]_{y^{2} / 3}^{2 y} \, d y = \int_{0}^{6} \left( 2y - \frac{y^{2}}{3} \right) \, d y \]Simplify the integrand:\[ \int_{0}^{6} \left( \frac{6y}{3} - \frac{y^{2}}{3} \right) \, d y = \int_{0}^{6} \left( 2y - \frac{y^{2}}{3} \right) \, d y = \int_{0}^{6} \left( 2y - \frac{y^{2}}{3} \right) \, d y \]Calculate the integral:\[ \left[ y^2 - \frac{y^3}{9} \right]_0^6 = \left( 36 - \frac{216}{9} \right) - 0 = 36 - 24 = 12 \].

Step 5: Conclusion

Therefore, the area of the region bounded by the curves \( x = \frac{y^{2}}{3} \) and \( x = 2y \) from \( y = 0 \) to \( y = 6 \) is 12 square units.

Key Concepts

Area between Curves

The area between curves is a fundamental concept in calculus, referring to the region enclosed between two distinct curves on a graph. When computing this area using double integrals, we set the bounds of integration to the limits defined by the intersection points of the curves. In our original exercise, the curves given by the equations \( x = \frac{y^2}{3} \) and \( x = 2y \) form the limits of integration and, thus, determine the region whose area we wish to calculate.
To find the area, we need to integrate with respect to both variables, typically performing the inner integral first. For our problem, since \( x \) varies between the given curves, the inner integral is with respect to \( x \), and the outer integral is with respect to \( y \).
This setup allows us to neatly calculate the total area by subtracting the integral of one curve from the other, providing a clear understanding of the space encapsulated between the curves.

Intersection Points

Intersection points are the locations where two or more curves meet on a graph. Finding these points is crucial for determining the bounds of integration when calculating the area between curves. To find the intersection points of the curves \( x = \frac{y^2}{3} \) and \( x = 2y \), we set the equations equal to each other and solve for \( y \).
For the original exercise, solving \( \frac{y^2}{3} = 2y \) results in the equation \( y(y - 6) = 0 \), giving us \( y = 0 \) or \( y = 6 \). Substituting these \( y \)-values back into either equation gives the points \((0, 0)\) and \((12, 6)\) as the intersection points.
These intersection points define the limits for the variable \( y \), crucial for setting up the correct boundaries for our double integrals.

Sketching Regions

Sketching the region of integration helps to visualize the calculated area and the bounding curves. By drawing the curves on a coordinate plane, we see their paths and the space between them. This step also aids in identifying any potential mistakes in calculations when setting up integrals.
For our original problem, sketch the curves \( x = \frac{y^2}{3} \) and \( x = 2y \). The curve \( x = \frac{y^2}{3} \) appears as a parabola opening to the right, while \( x = 2y \) is a straight line through the origin. The intersection at points \((0, 0)\) and \((12, 6)\) encloses the region of interest.

• Draw the parabola starting from the origin, bending outwards.
• Draw the line passing through the origin achieving the intersection with the parabola at \((12, 6)\).
• Highlight the area between the curves from \( y = 0 \) to \( y = 6\).

This visual representation solidifies understanding and ensures that the integration setup captures the intended area.

Definite Integration

Definite integration involves calculating the integral of a function within a specific set of bounds, providing a precise numerical result as opposed to indefinite integration which leaves a constant of integration.
In the context of finding area using double integrals, definite integrals allow for the exact calculation of the size of the region. For the given exercise, after setting up our integral from earlier steps, we integrate first with respect to \( x \), followed by \( y \).
This results in evaluating:\[\int_{0}^{6} \left[ x \right]_{y^{2}/3}^{2y} \, dy = \int_{0}^{6} \left( 2y - \frac{y^2}{3} \right) \, dy\]Solving this, we find:\[\left[ y^2 - \frac{y^3}{9} \right]\bigg|_0^6 = 12\]Thus, the area bounded by these curves is 12 square units. Definite integration simplifies complex functions into tangible values, offering a complete understanding of areas on the xy-plane.