Well, sure. Otherwise, how do you know the limits of integration?
I'm sure you found that the curves intersect at (1,5) and (9,-3)
So, using horizontal strips of width dy, the area is
∫[-3,5] (6-y) - (y-3)^2/4 dy = 64/3
You could use vertical strips of width dx, but then you have to allow for the two branches of the parabola, and divide the region where the boundary changes:
∫[0,1] (3+2√x)-(3-2√x) dx + ∫[1,9] (6-x)-(3-2√x) dx
Hi, not sure how to set this integral up for my Calculus II homework.
"Given a graph that contains a parabola and a line, the equation of the parabola is x=(y-3)²/4 and the equation of the line is y=6-x. Find the area of the shaded region".
Are we supposed to find the points of intersection between these two equations?
1 answer
0 answers