y = x^2
y = 6 - x
Find the area of the region by integrating (a) with respect to x and (b) with respect to y.
To find the area of the region bounded by the two curves, we need to find the points of intersection first. Let's solve the two equations:
1) Set the two equations equal to each other:
x^2 = 6 - x
2) Rearrange the equation into standard quadratic form:
x^2 + x - 6 = 0
3) Factorize the quadratic equation:
(x + 3)(x - 2) = 0
So, we have x + 3 = 0 or x - 2 = 0
4) Solve for x:
x = -3 or x = 2
We've found the x-coordinates of the points of intersection. Now, let's find the corresponding y-coordinates by substituting the x-values into one of the equations. Let's use the equation y = 6 - x:
For x = -3:
y = 6 - (-3) = 6 + 3 = 9
For x = 2:
y = 6 - 2 = 4
So, the points of intersection are (-3, 9) and (2, 4). Now, let's proceed to find the area using different methods:
(a) Area with respect to x:
To find the area with respect to x, we need to integrate the difference in y-values (y2 - y1) between the curves with respect to x, over the interval from x1 to x2.
The area can be calculated using the integral:
A = ∫[x1, x2] (y2 - y1) dx
In this case, the upper curve is y = x^2 and the lower curve is y = 6 - x.
So, the area can be calculated as:
A = ∫[-3, 2] [(x^2) - (6 - x)] dx
Simplifying the equation:
A = ∫[-3, 2] [x^2 + x - 6] dx
Evaluate the integral:
A = [(1/3)x^3 + (1/2)x^2 - 6x] |[-3, 2]
Plug in the limits of integration:
A = [(1/3)(2)^3 + (1/2)(2)^2 - 6(2)] - [(1/3)(-3)^3 + (1/2)(-3)^2 - 6(-3)]
Simplify and calculate the values:
A = [(8/3) + 2 - 12] - [(-27/3) + (9/2) + 18]
A = [(8/3) - 10] - [(-27/3) + (9/2) + 18]
A = (8/3 - 30/3) - (-27/3 + 9/2 + 18)
A = (-22/3) - (1/3)
A = -23/3
Therefore, the area with respect to x is -23/3 square units.
(b) Area with respect to y:
To find the area with respect to y, we need to integrate the difference in x-values (x2 - x1) between the curves with respect to y, over the interval from y1 to y2.
The area can be calculated using the integral:
A = ∫[y1, y2] (x2 - x1) dy
In this case, the upper curve is y = x^2 and the lower curve is y = 6 - x.
So, the area can be calculated as:
A = ∫[4, 9] [(√y) - (6 - √y)] dy
Simplifying the equation:
A = ∫[4, 9] [(2√y) - 6] dy
Evaluate the integral:
A = [(2/3)y^(3/2) - 6y] |[4, 9]
Plug in the limits of integration:
A = [(2/3)(9)^(3/2) - 6(9)] - [(2/3)(4)^(3/2) - 6(4)]
Simplify and calculate the values:
A = [(2/3)(27) - 54] - [(2/3)(8) - 24]
A = (18 - 54) - (16/3 - 24)
A = -36 - (-16/3)
A = -108/3 + 16/3
A = -92/3
Therefore, the area with respect to y is -92/3 square units.
Note: The negative sign indicates that the area is below the x-axis, which doesn't have a physical interpretation in this context. Thus, you may want to take the absolute value of the calculated area to get the positive value.