Sketch the region enclosed by the given curves. Decide whether to integrate with respect to x or y. Draw a typical approximating rectangle and label ts height and width. Then find the area of the region.
y=x^2
y^2=x
It is easy to see that the two curves intersect at (0,0) and (1,1)
I will integrate with respect to x
area = [integral] (x^(1/2) - x^2) dx from 0 to 1
= [(2/3)x^(3/2) - (1/3)x^3] from 0 to 1
= 2/3 - 1/3 - 0
= 1/3
To sketch the region enclosed by the given curves and find its area, we first need to analyze the equations and determine the boundaries of the region.
The first curve, y = x^2, represents a parabolic shape where y increases as x increases. The second curve, y^2 = x, represents a sideways parabola that opens to the right.
To find the boundaries of the region, we can solve the equations simultaneously. We have two possibilities: solving for y in terms of x or solving for x in terms of y.
First, we'll solve for y in terms of x:
From the second equation, y^2 = x, we take the square root of both sides to get y = √x.
Next, we'll solve for x in terms of y:
From the first equation, y = x^2, we can take the square root of both sides to get √y = x. However, since we're dealing with the region enclosed, we are only interested in the portion positive for x, so x = √y is our relevant equation.
Now, we can plot the two equations y = x^2 and x = √y on a graph to get an idea of what the region looks like:
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| \ (x = √y)
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The region is enclosed between the curves and is limited by the points of intersection, where y = x^2 and x = √y intersect. These intersection points can be found by setting the two equations equal to each other:
x^2 = √y
Raising both sides to the power of 4 to eliminate the square root:
(x^2)^2 = (√y)^4
x^4 = y^2
Now, we have two equations with the same variable:
x^4 = y^2
y^2 = x
To find the intersection points, we can solve these equations simultaneously by substituting the second equation into the first equation:
x^4 = (x)^2
Solving for x:
x^4 - x^2 = 0
x^2(x^2 - 1) = 0
This equation gives us two solutions:
x^2 = 0, which implies x = 0
x^2 - 1 = 0, which implies x = ±1
Now that we have the x-coordinates of the intersection points, we can determine the corresponding y-values by substituting the x-values into the equation y = x^2:
y = (0)^2 = 0
y = (±1)^2 = 1
So, the intersection points are (0, 0) and (±1, 1).
Now that we have the boundaries of the region on the xy-plane, we can integrate to find its area. Considering the shape of the region, it is more convenient to integrate with respect to y.
The bounded region lies between the two curves, y = x^2 and x = √y, from y = 0 to y = 1.
To find the area, we integrate the difference between the curves with respect to y:
Area = ∫[from 0 to 1] [(√y) - (y^2)] dy
Using the limits of integration, we can evaluate this integral to find the area of the region.
Now, let's draw a typical approximating rectangle to understand the dimensions involved:
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|wwwwwwwww|
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h | |
|________|
In the rectangle, the width (w) will be the difference between the x-coordinate values of the curves (√y - y^2), and the height (h) will be the difference between the upper and lower limits of integration.
Finally, after evaluating the integral and performing the necessary calculations, we will obtain the area of the region enclosed by the given curves.