Math Problem Statement

ketch the region enclosed by x + y^2 = 20 and x + y = 20 . Decide whether to integrate with respect to x or y , and then find the area of the region.

The area is:

Solution

To solve the problem of finding the area of the region enclosed by the curves x+y2=20x + y^2 = 20 and x+y=20x + y = 20, we can follow these steps:

1. Sketch the Curves

First, let's rewrite the equations to better understand the curves:

  1. Equation 1: x+y2=20x + y^2 = 20

    • Solving for xx: x=20y2x = 20 - y^2
    • This represents a parabola opening to the left with vertex at (20,0)(20, 0).
  2. Equation 2: x+y=20x + y = 20

    • Solving for xx: x=20yx = 20 - y
    • This represents a straight line with a slope of 1-1.

2. Find the Points of Intersection

To find the points of intersection between the parabola and the line, set the equations equal to each other:

20y2=20y20 - y^2 = 20 - y

Simplifying, we get:

y2=yy^2 = y y(y1)=0y(y - 1) = 0

So, y=0y = 0 or y=1y = 1.

Substituting these values of yy back into one of the original equations to find xx:

  • For y=0y = 0: x=2002=20x = 20 - 0^2 = 20.
  • For y=1y = 1: x=201=19x = 20 - 1 = 19.

Thus, the points of intersection are (20,0)(20, 0) and (19,1)(19, 1).

3. Decide the Method of Integration

Given the nature of the equations (especially the parabola opening horizontally), it is easier to integrate with respect to yy.

4. Set Up the Integral

The area AA can be found by integrating the difference between the two functions from y=0y = 0 to y=1y = 1:

A=01[(20y)(20y2)]dyA = \int_{0}^{1} \left[(20 - y) - (20 - y^2)\right] dy

Simplifying the integrand:

A=01(y+y2)dyA = \int_{0}^{1} \left(-y + y^2\right) dy

5. Evaluate the Integral

Now, integrate:

A=01(y+y2)dy=[y22+y33]01A = \int_{0}^{1} \left(-y + y^2\right) dy = \left[-\frac{y^2}{2} + \frac{y^3}{3}\right]_{0}^{1}

Substituting the limits:

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

A=[12+13]=36+26=16A = \left[-\frac{1}{2} + \frac{1}{3}\right] = -\frac{3}{6} + \frac{2}{6} = -\frac{1}{6}

Since we are finding the area, we take the absolute value:

A=16A = \frac{1}{6}

Final Answer:

The area of the region enclosed by the curves is 16\boxed{\frac{1}{6}}.


Would you like more details on any part of this process, or do you have any questions?

Related Questions:

  1. How would you sketch the graph of x+y2=20x + y^2 = 20?
  2. What happens if you integrate with respect to xx instead?
  3. How do you find the intersection points of other types of curves?
  4. How does the direction in which a parabola opens affect the integration method?
  5. Can you explain the significance of taking the absolute value when calculating area?

Tip:

When deciding whether to integrate with respect to xx or yy, choose the variable that simplifies the integral and makes the bounds easier to determine.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Integration
Area Calculation
Quadratic Equations

Formulas

Integration by Substitution

Theorems

-

Suitable Grade Level

Grades 11-12