Integrating an area bounded by three lines/curves Given the three equations: 
$$y_1 = 25 - x^2$$
$$y_2 = 25 - \frac{25}{3}x$$ 
$$y_3 = 9x - 27$$
I need to find the area bounded by the graphs of each equation  in the first quadrant. The graph is shown here: 
Graph
Points of Intersection are: (0,25) [y1 & y2] ; (3,0) [y2 & y3] ; (4,9) [y1 & y3]
Thus, the area between these points is the area that needs to be calculated.
If anyone could give some type of direction on how to set up the integral(s), that would be much appreciated. I'm guessing it needs to be split into two or three integrals somehow, but not sure where to start.
 A: As you said, we have $(0, 25)$, $(3, 0)$, and $(4, 9)$ as our intersection points. Therefore, we need to look at the regions of area in between those intersections points.
Between $x=0$ and $x=3$, the area is between the blue curve, $y=25-x^2$, and the purple curve, $y=25-\frac{25x}{3}$. Thus, we have the following integral:
$$\int_0^3 \left(25-x^2-\left(25-\frac{25x}{3}\right)\right)dx$$
Between $x=3$ and $x=4$, the area is between the blue curve, $y=25-x^2$, and the red curve, $y=9(x-3)$. Thus, we have the following integral:
$$\int_3^4 (25-x^2-(9(x-3)))dx$$
Add the two integrals together to get the total area.
A: This looks like a homework problem, and I think there might even be a tag for such problems.
Regardless, from the graph you can see that you can construct the area with two integrals (vertical or horizontal). I'll illustrate vertical.
The total area between functions $f(x)$ and $g(x)$ where $f$ is above $g$, may be written as
$A = \int_a^b (f(x) - g(x) )dx$
Since $y_1$ is the top function for both segments, and the integral must be split at $x=3$, we have
$A = \int_0^3 (y_1 - y_2) dx + \int_3^4 (y_1 - y_3 ) dx$
Then plug in your expressions for $y_1,y_2,y_3$ and integrate.
Substituting, we have
$A = \int_0^3 (\frac{25}{3} x - x^2) dx + \int_3^4 (52 - 9x - x^2 ) dx$
Performing this integral (which is straight forward) yields $A= \frac{110}{3}$.
A: Well, according to your figure, you need to write the integral as a sum of two integrals.
$y = 25 - x^2\rightarrow x=\sqrt{25-y}$ (In the first quadrant).
$y = 25 - \frac{25}{3}x\rightarrow x= 3 - \frac{3}{25}y$.
$y = 9x - 27\rightarrow x=3+ \frac{1}{9}y$.
Now, for $0\leq y\leq 4$, we have $3 - \frac{3}{25}y\leq x\leq 3+ \frac{1}{9}y$, which is the bounds for the first integral.
Next, for $4\leq y\leq 25$, we have $3 - \frac{3}{25}y\leq x\leq 3 - \frac{3}{25}y$, which is the bounds for the second integral.
Your answer is the summation of these two integrals. Please note that I had a double integral in my mind.
