# Thread: area between the curves

1. ## area between the curves

so I have a question here, in a line of questions, where i am supposed to figure out of a) i need to integrate with respect to x or y, and b)find the area of the region.

the problem is 4x+y^2=12, x=y. so i figure ill integrate with respect to y, and i move the x over to the other side in the first problem, move the 12 to the left side, then divide everything by 4 to come to 4-(y^2)/4=x now when i sketch this, one of the intercepts is to the left of the y axis, and one is to the right.

I really don't know where to go at this point.

2. Your first mistake is that you made an arithmetic error in solving $4x+ y^2= 12$ for x. 12/4= 3, not 4! $x= 3- (1/4)y^2$ which is a parabola with x- axis as axis of symmmetry, vertex at (-3, 0), and opening to the right. The line x= y is always the left boundary and the parabola the right boundary of the area they define so you want to subtract the x value of the line from the x value of the parabola to get a positive value: $y- (3- (1/4)y^2)$.

Yes, one of the points of intersection is to the left of the y-axis and one to the right but that is not really relevant. What are the y coordinates of those points. You want to integrate $\int_{y_1}^{y_2} (y- 3+ (1/4)y^2) dy$ where $y_1$ is the y coordinate of the lower point and $y_2$ is the y coordinate of the higher point.

3. how does it open right? its -(1/4)y^2 isnt it? sorry bout the 4, i was running out the door this morning, and I did have 3 :P also, wouldnt I integrate (3-1/4y^2)-y?