Question

Sorry for the following mess; I don't know any tagging code. Question: find the area between the two lines, but with respect to y. y = 4 - x^2 and y - x - 2 = 0. So far, I've tried to rewrite them in terms of x: x = sqrt(4-y) and x = y - 2. I then set them equal to each other to find the intersections (I came up with y = 0, 3). Here is where I'm completely lost. I don't know which line to subtract from which, and either way I tried comes out incorrect. for "d" I used 3, and for "c" I used 0. [ (y^2)/2 - 2y + (2/3)(4-y)^(3/2) ]. The answer in the back of book is 9/2, if that helps any.

Answer 1

\[\int\limits_{0}^{3}\left( y-2-\sqrt{4-y} \right)dy \rightarrow \left( \frac{ y^2 }{ 2 }-2y+\frac{ 2\left( 4-y \right)^\frac{ 3 }{ 2 } }{ 3 } \right)\]

Answer 2

Draw the curve and lines, and you should be OK.

Answer 3

\[Area = \int\limits_{a}^{b} 4-x ^{2} -(x+2) dx\]

Answer 4

You have calculated the intersection of the lines wrong, by the way.
Check again and insert into the a and b above

Answer 5

I need it with respect to "y." And, not for nothing, but I've been staring at the graph all afternoon. I'm thinking now that I should be using 4 for d and still 0 for c, but I don't know whether to subtract the parabola (in terms of x) from the line (in terms of x) or the other way around.

Answer 6

I eventually got it. For anyone that reads this or has a similar question, here is what worked: from 4 to 3 make sure to multiply the area by 2 (because when I converted the tip of the parabola to terms of x, I only took the "+" [i.e. half]). 2nd, and this is the most embarrassing of all, if you look at my equation above you will notice that I am ADDING(!) the curve I really wanted to subtract. Fixing those two things yielded the correct answer.