Sketch the region enclosed by the curves x= 49-y^2 and x = y^2 - 49. Decide whether to integrate with respect to x or y. Then find the area of the region.
sketching should have been no problem
both have axes on the x-axis,
one has vertex (-49,0) and opens to the right
the other has vertex (49,0) and opens to the left
There is perfect symmetry and they meet at (0,7) and (0,-7)
I would take horizontal slices, thus integrate with respect to y
Also go from y = 0 to y = 7 and double that answer
area = 2(integral) (49 - y^2 - (y^2-49)) dy from 0 to 7
= 2(integral) (98 - 2y^2) dy from 0 to 7
= 2 [98y - (2/3)y^3] from 0 to 7
= 2(686 - (2/3)(343) - 0 )
= 2(1372/3)
= 2744/3 or 914 2/3 units^2
confirmation:
http://www.wolframalpha.com/input/?i=area+x%3D+49-y%5E2+%2C+x+%3D+y%5E2+-+49
I don't know why Wolfram turned the graph sideways, notice the y-axis is horizontal and the x-axis is vertical
Well, let's start by visualizing this comical circus of curves. We have one curve, x = 49-y^2, which is like a parabola that opens downward. And then we have another curve, x = y^2 - 49, which is like a parabola that opens upward. Now, let's roll out the circus tent and see what we're dealing with here!
To determine the region enclosed by these curves, we need to find the points where they intersect. Equating the two equations, we get y^2 - 49 = 49 - y^2. So, adding y^2 to both sides gives us 2y^2 = 98, which simplifies to y^2 = 49. Taking the square root, we get y = ±7.
So, we have two points of intersection, where y = 7 and y = -7. Now, let's find out whether to integrate with respect to x or y. To do this, let's consider the shape of our curves.
The curve x = 49 - y^2 looks like a sideways parabola, and the curve x = y^2 - 49 also looks like a sideways parabola. When we sketch them together, we see that they intersect at y = ±7 and form a closed figure. This means our region is bounded by y = -7 and y = 7, and the curves themselves.
Now, it's party time! We're going to integrate with respect to y since the given curves are expressed as functions of y. So our integral will be ∫[from -7 to 7] (x2 - (49 - x)2) dy.
Before we dive into calculations, let's simplify a bit. Expanding (49 - x)2 gives us (49 - x)^2 = 2401 - 98x + x^2.
Now, let's integrate! ∫[from -7 to 7] (x2 - (2401 - 98x + x^2)) dy.
Simplifying further, we get ∫[from -7 to 7] (2401 - 98x + 2x^2) dy.
Now, integrating term by term, we get ∫[from -7 to 7] (2401y - 98xy + 2xy^2) dy.
Evaluating the integral, we find the area enclosed by the curves to be 2,401(14) - 98x(0) + 2x(392/3) = 33,614 + 0 + 784/3 = 33,614.26
So, the area of the region is approximately 33,614.26 square units. Enjoy the show, and remember, always look for the humor in math!
To sketch the region enclosed by the curves x = 49 - y^2 and x = y^2 - 49, let's first find the points of intersection between these two curves.
Setting x = 49 - y^2 equal to x = y^2 - 49, we have:
49 - y^2 = y^2 - 49
Rearranging the equation:
y^2 + y^2 = 98
2y^2 = 98
y^2 = 49
Taking the square root of both sides:
y = ±7
Thus, the two curves intersect at (7, 7) and (-7, -7).
Now, let's analyze the behavior of the curves. The curve x = 49 - y^2 is a downward-opening parabola that opens leftward and is symmetric about the y-axis. The curve x = y^2 - 49 is an upward-opening parabola that opens rightward and is also symmetric about the y-axis.
Since the region is enclosed between these two curves, we need to calculate the area by integrating with respect to y.
Using the formula for the area between two curves with respect to y, the area (A) can be calculated as follows:
A = ∫[from y = -7 to y = 7] [(y^2 - 49) - (49 - y^2)] dy
Simplifying the integrand:
A = ∫[from y = -7 to y = 7] [2y^2 - 98] dy
Integrating term by term:
A = (2/3)y^3 - 98y evaluated from y = -7 to y = 7
Substituting the limits of integration:
A = [(2/3)(7)^3 - 98(7)] - [(2/3)(-7)^3 - 98(-7)]
Evaluating the expression:
A = [(2/3)(343) - 686] - [(2/3)(-343) + 686]
A = [(686/3) - 686] - [(-686/3) + 686]
A = [686/3 - 686] + [686/3 - 686]
Simplifying:
A = 686/3 - 686 + 686/3 - 686
A = 2(686/3) - 1372
A = 1372/3 - 1372
A = 1372/3 - 4116/3
A = -2744/3
Therefore, the area of the region enclosed by the curves x = 49 - y^2 and x = y^2 - 49 is -2744/3 square units.
To sketch the region enclosed by the curves x = 49 - y^2 and x = y^2 - 49, let's first analyze the curves individually.
1. Curve 1: x = 49 - y^2
This is a downward opening parabola with its vertex at (49, 0) and its axis of symmetry aligned with the y-axis. It opens downward because the squared term is negative, and it is shifted 49 units to the right.
2. Curve 2: x = y^2 - 49
This is an upward opening parabola with its vertex at (-49, 0) and its axis of symmetry aligned with the y-axis. It opens upward because the squared term is positive, and it is shifted 49 units to the left.
To sketch the region, we need to ascertain the points of intersection between the two curves. Setting the two equations equal to each other:
49 - y^2 = y^2 - 49
Rearranging:
2y^2 = 98
Dividing both sides by 2:
y^2 = 49
Taking the square root:
y = ± 7
So, the two curves intersect at y = 7 and y = -7.
Now, to decide whether to integrate with respect to x or y to find the area of the region, we need to determine whether the region is more easily expressed in terms of x or y coordinates.
Looking at the curves, it is clear that the region is vertically bounded. To calculate the area, we need to integrate with respect to y. Therefore, we will integrate using the formula:
A = ∫[a,b] (x₂ - x₁) dy
where a and b are the y-values at which the curves intersect.
Now let's find the area step-by-step:
1. Determine the bounds of integration:
Since the two curves intersect at y = 7 and y = -7, these will be our bounds of integration: a = -7 and b = 7.
2. Identify the functions that represent the top and bottom of the region:
x = y^2 - 49 will be the top curve, and x = 49 - y^2 will be the bottom curve.
3. Determine the limits for x in terms of y:
At y = -7, the bounds of x are x = -7^2 - 49 = -98, and x = 49 - (-7)^2 = 0.
At y = 7, the bounds of x are x = 7^2 - 49 = 0, and x = 49 - 7^2 = 0.
4. Set up and evaluate the definite integral:
The area A will be given by the definite integral:
A = ∫[-7,7] ((y^2 - 49) - (49 - y^2)) dy
Simplifying the integrand:
A = ∫[-7,7] (2y^2 - 98) dy
Integrating:
A = [(2/3)y^3 - 98y] evaluated from y = -7 to y = 7
Evaluating at the limits:
A = [(2/3)(7^3) - 98(7)] - [(2/3)(-7^3) - 98(-7)]
Simplifying:
A = [(686/3) - 686] - [(-686/3) - 686]
Calculating:
A = (686/3) - 686 + (686/3) + 686
Simplifying:
A = 2(686/3) = 457.33
Therefore, the area of the region enclosed by the curves x = 49 - y^2 and x = y^2 - 49 is approximately 457.33 square units.