Sketch the region enclosed by x+y^2=42 and x+y=0.
Decide whether to integrate with respect to x or y, and then find the area of the region.
To sketch the region enclosed by the equations x + y^2 = 42 and x + y = 0, we can start by solving the two equations individually.
1) Solve x + y^2 = 42 for y:
Rearrange the equation to isolate y^2:
y^2 = 42 - x
y = ± √(42 - x)
2) Solve x + y = 0 for y:
y = -x
Now, plot the graphs of y = ± √(42 - x) and y = -x on a coordinate axis to visualize the region enclosed by these curves.
The graph of y = ±√(42 - x) is a pair of parabolas symmetric about the x-axis, while the graph of y = -x is a straight line that passes through the origin and has a negative slope.
To determine the region enclosed, we need to find the points of intersection between these curves. Setting the expressions for y equal to each other, we have:
±√(42 - x) = -x
Now, we can square both sides to simplify the equation:
42 - x = x^2
Rearrange this equation:
x^2 + x - 42 = 0
Solve for x using factoring, quadratic formula, or other applicable methods. The solutions will give us the x-coordinates of the points of intersection between the curves.
Once we have the x-coordinates, we can substitute them back into either equation to find the corresponding y-values.
Next, we can determine whether to integrate with respect to x or y to find the area of the region. In this case, since the equations are defined in terms of x and y, it is simpler to integrate with respect to y.
The limits of integration for y can be determined by the range of y-values within the enclosed region. This can be found by analyzing the point(s) of intersection and the behavior of the curves.
After determining the limits of integration, we can set up the integral to find the area:
A = ∫[lower limit, upper limit] (y2 - y1) dx
= ∫[lower limit, upper limit] (√(42 - x) - (-x)) dx
= ∫[lower limit, upper limit] (√(42 - x) + x) dx
Evaluate this integral to find the area of the region enclosed by x + y^2 = 42 and x + y = 0.
Note: The steps provided are a general guide to sketching and finding the area of the region enclosed by the given equations. Specific values and calculations will vary depending on the nature of the equations and points of intersection.
To sketch the region enclosed by the equations x+y^2=42 and x+y=0, we can start by analyzing the equations separately.
1. x + y^2 = 42:
This equation represents a sideways "U-shaped" curve called a parabola. To graph it, we can rearrange the equation as follows:
y^2 = 42 - x
y = ±√(42-x)
By plugging different values of x into the equation and calculating the corresponding y values, we can plot points and sketch the parabola.
2. x + y = 0:
This equation represents a straight line passing through the origin with a slope of -1. By rearranging the equation, we have:
y = -x
So, this line intersects the x-axis at the point (1, -1) and the y-axis at the point (0, 0).
Now, to find the region enclosed by these equations, we need to identify the points where they intersect. The equations are x + y^2 = 42 and x + y = 0.
By substituting -x for y in the first equation, we get:
x + (-x)^2 = 42
x + x^2 = 42
x^2 + x - 42 = 0
Solving this quadratic equation, we find that it factors as (x + 7)(x - 6) = 0. Therefore, x can be either -7 or 6.
When x = -7:
From the second equation, y = -x = -(-7) = 7.
So, one intersection point is (-7, 7).
When x = 6:
From the second equation, y = -x = -(6) = -6.
So, the other intersection point is (6, -6).
Now, let's plot these points and sketch the region enclosed by the two equations:
(6, -6) (-7, 7)
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Origin (0,0)
(x-axis)
The region enclosed by the curves is the shaded area between the parabola and the line.
To find the area of this region, we can integrate with respect to either x or y. In this case, it is simpler to integrate with respect to x.
Since the line y = -x is below the parabola y^2 = 42 - x in this region, the limits of integration for x will be from -7 to 6.
So, the area can be calculated by integrating the difference between the curves with respect to x:
Area = ∫[from -7 to 6] [√(42-x) - (-x)] dx
By evaluating this integral, you can find the area of the region enclosed by the given equations.