Problem 13 - Thomas Calculus

Question

The integrals and sums of integrals in Exercises \(13 - 18\) give the areas of regions in the \(x y\) -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 units squared.

Step 1: Identify the region of integration

The given integral is \( \int _ { 0 } ^ { 6 } \int _ { y ^ { 2 } / 3 } ^ { 2 y } d x d y \). The limits for \( x \) range from \( \frac{y^2}{3} \) to \( 2y \) and for \( y \) from 0 to 6. This suggests that \( x \) is bounded by the curves defined by \( x = \frac{y^2}{3} \) and \( x = 2y \).

Step 2: Sketch the region and bounding curves

To understand the region, sketch the curves \( x = \frac{y^2}{3} \) and \( x = 2y \) on the coordinate plane. These curves will form a closed region extending vertically from \( y = 0 \) to \( y = 6 \). The curve \( x = \frac{y^2}{3} \) is a parabola opening to the right, while \( x = 2y \) is a straight line.

Step 3: Find intersection points of curves

Set the equations \( \frac{y^2}{3} = 2y \) to find the points of intersection. Simplifying gives \( y^2 = 6y \), or \( y(y - 6) = 0 \). Thus, \( y = 0 \) or \( y = 6 \). Substitute these values back into either of the equations to find \( x \) values: when \( y = 0 \), \( x = 0 \); when \( y = 6 \), \( x = 12 \). Intersection points are \((0, 0)\) and \((12, 6)\).

Step 4: Set up the integral for the area

The area of the region between these curves and within the given bounds can be found by evaluating the double integral \( \int _ { 0 } ^ { 6 } \int _ { y ^ { 2 } / 3 } ^ { 2 y } d x d y \). We will first integrate with respect to \( x \) and then with respect to \( y \).

Step 5: Integrate with respect to x

For the inner integral, evaluate \( \int _ { y^2/3 } ^ { 2y } d x = [x]_{y^2/3}^{2y} = 2y - \frac{y^2}{3} \). This gives the expression for the area between the two \( x \)-curves at any point \( y \).

Step 6: Integrate with respect to y

Now perform the outer integration over \( y \): \( \int _ { 0 } ^ { 6 } \left( 2y - \frac{y^2}{3} \right) d y \). The antiderivative is \( \int (2y - \frac{y^2}{3}) d y = [y^2 - \frac{y^3}{9}]_{0}^{6} \).

Step 7: Calculate the definite integral

Calculate \( [y^2 - \frac{y^3}{9}]_{0}^{6} \). Substitute \( y = 6 \) to get \( 36 - \frac{216}{9} \) and \( y = 0 \) to get 0. Thus, the result is \( 36 - 24 = 12 \).

Step 8: Conclude with the area

The area of the region defined by the curves and limits is 12.

Key Concepts

Area Calculation

To find the area of a region in the coordinate plane, one effective method is using double integrals. Double integrals work by summing up infinitely small areas across a defined region. This method is particularly helpful when the region is bounded by curves and lines, as in this exercise.
To calculate the area using the double integral \( \int_{0}^{6} \int_{y^2/3}^{2y} dx\, dy \), we first integrate with respect to \( x \) and then with respect to \( y \). The inner integration with respect to \( x \) gives the expression for the area between the two curves at any point \( y \).
Once you have this expression, you perform the outer integration with respect to \( y \), from 0 to 6, to sum up these small slices of area for the entire vertical span. This results in the total area. For this specific region, the area is calculated to be 12 square units.

Region of Integration

The region of integration is crucial in setting up your double integrals. It refers to the area over which you are integrating. Understanding this region can guide the setup of the limits in your integral.
In this exercise, the given integral \( \int_{0}^{6} \int_{y^2/3}^{2y} dx\, dy \) tells us:

• The variable \( x \) is bounded by the curves \( x = \frac{y^2}{3} \) and \( x = 2y \).
• The variable \( y \) spans from 0 to 6.

Consequently, the region of integration is a planar figure shaped between these curves. This setup ensures you calculate the area only within the defined bounds, allowing the double integral to account for changes between the curves across \( y \).

Intersection Points

Intersection points are the spots where two curves meet or cross each other in the plane. Finding these is important because they usually determine the boundaries for calculating areas using integrals.
To find these points for the curves \( x = \frac{y^2}{3} \) and \( x = 2y \), we set the equations equal: \( \frac{y^2}{3} = 2y \). Solving for \( y \), we find \( y(y-6) = 0 \), yielding \( y = 0 \) and \( y = 6 \).
Subsequently, plug these \( y \)-values back into either equation to determine the corresponding \( x \)-values. At \( y = 0 \), \( x = 0 \); at \( y = 6 \), \( x = 12 \). Thus, the intersection points are \( (0, 0) \) and \( (12, 6) \). These coordinates are essential as they dictate the start and end of the region over which you integrate.

Sketching Graphs

Sketching the graphs of the given curves helps visualize the region of integration, making it easier to set up the double integral. In this exercise, we work with the curves \( x = \frac{y^2}{3} \) and \( x = 2y \).
The curve \( x = \frac{y^2}{3} \) is a parabolic shape opening to the right, and \( x = 2y \) is a straight, slanted line. These intersect at the points found earlier, \( (0, 0) \) and \( (12, 6) \).
By plotting these curves on a coordinate axis, you can clearly see the closed region they form between \( y = 0 \) and \( y = 6 \). This sketch acts as a map, aiding in correctly setting bounds for your integral. It ensures that the area calculated encompasses the full region of interest and not beyond. Visualization complements the analytical setup, providing a more intuitive grasp of the problem.