consider the area between the graphs x+3y=1 and x+9=y^2. this area can be computed in two different ways using integrals. Please integral both ways both respect to x and to y. ANd please show your work and thank you
To find the area between the graphs x+3y=1 and x+9=y^2, we can calculate it using two different methods: integrating with respect to x and integrating with respect to y. Let's work through both methods step by step.
Method 1: Integrating with Respect to x
To integrate with respect to x, we need to express the equations in terms of x.
The first equation, x+3y=1, can be rewritten as y = (1-x)/3.
The second equation, x+9=y^2, can be rearranged as x = y^2 - 9.
Now, let's find the x-limits of the area. To do this, we set the two equations equal to each other:
(1-x)/3 = y^2 - 9
Rearranging this equation, we get:
x = 1 - 3y^2 + 27
x = 28 - 3y^2
Now, let's find the limits for y. To do this, we'll set both equations equal to zero:
(1-x)/3 = 0
1 - x = 0
x = 1
x + 9 = y^2
y^2 = x + 9
Thus, the limits for y are:
Lower limit: y = √(x + 9)
Upper limit: y = (1-x)/3
Next, we can set up the integral to find the area:
A = ∫[Lower limit, Upper limit] (28 - 3y^2) - √(x + 9) dx
To evaluate this integral, we need to simplify the expression first:
A = ∫[Lower limit, Upper limit] 28 - 3y^2 - √(x + 9) dx
Now, integrate the expression with respect to x and evaluate the definite integral using the respective limits.
Method 2: Integrating with Respect to y
To integrate with respect to y, we need to express the equations in terms of y.
The first equation, x+3y=1, can be rewritten as x = 1 - 3y.
The second equation, x+9=y^2, can be rearranged as x = y^2 - 9.
Now, let's find the y-limits of the area. To do this, we set the two equations equal to each other:
1 - 3y = y^2 - 9
Rearranging this equation, we get:
y^2 + 3y - 10 = 0
This equation can be factored as:
(y + 5)(y - 2) = 0
Solving for y, we find two solutions:
y = -5 and y = 2
Thus, the limits for y are:
Lower limit: y = -5
Upper limit: y = 2
Next, we can set up the integral to find the area:
A = ∫[Lower limit, Upper limit] [(1 - 3y) - (y^2 - 9)] dy
To evaluate this integral, we need to simplify the expression first:
A = ∫[Lower limit, Upper limit] 10 - y^2 + 3y dy
Now, integrate the expression with respect to y and evaluate the definite integral using the respective limits.
Remember to substitute the limits into the integrals based on the calculations to find the area of the given region.