Asked by Sky
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?
"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?
Answers
Answered by
oobleck
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
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
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.