Area bounded by two curves

[NYC]

ok, I need to find the area of the region bounded by x=6-y^2 and x=y^2-2

to get the answer, I switched the x's and y's and got 64/3 ; however, I'd like to know how it would be done otherwise (i.e. the correct way) I took the integral of the difference between the upper function and lower function over the area bounded by the curves (between -2 and 2 after changing x to y and vice versa) Help would be greatly appreciated

 

[galactus]

If you solve for y and set up the integrals:

\(\displaystyle 2(\int_2^{6}\sqrt{6-x}dx+\int_{-2}^{2}\sqrt{x+2}dx)=\frac{64}{3}\)

 

[pka]

NYC: Integrate the difference of the ‘right most’ and ‘left most’ curves.

\(\displaystyle \L
\int_{ - 2}^2 {\left[ {\left( {6 - y^2 } \right) - \left( {y^2 - 2} \right)} \right]dy}\)

Note the points of intersection are (−2,2) & (2,2).

 

[galactus]

Ain't that something, pka. I integrated this and got 32/3. It should work out the

same, shouldn't it?.

Oh, I see. I forgot to integrate what is 'south of the x-axis'.

Your 'y' method is better than my radicals. Nonetheless, I thought they should be

the same answer.

 

[pka]

The graph with the MathCad equivalent program!

 

[NYC]

Thanks, I did it using the "right most and left most" curves and got the right answer. I also have a general question. Suppose you want to find the region bounded by two curves across an interval [a,b] You evaluate the integral for B, and get a positive number, and then evaluate for A, getting a negative. Do you discard the negative, or proceed with (assume F(x) is the antiderivative) F(b)-F(a)? My thinking is that you keep the negative, though my professor discarded it in one of his problems.

 

[pka]

“My thinking is that you keep the negative, though my professor discarded it in one of his problems.”
Actually it is just a matter of understanding the Riemann Approximating Sums.
Think about the area!
How would you approx the area using rectangles?