Area Between Curves

[InterserveVB]

Using calculus, I would like to find the area the region enclosed by x = 9 - y^2 and x = y^2 - 9

How would I do this?

Normally they would both be y =. I tried changing them into that form, but I kept getting weird answers. If I do it using the x= format and leaving the ys in my intergals do I use the x or the y values to limit my integral?

 

[stapel]

Normally they would both be y =.

Yes, normally they would both be in terms of "y=". Doing one of them in terms of "x = (something funky)" makes this a bit more awkward.

But try doing the graph. The "x = 9 - y²" is just a left-opening parabola (that is, a parabola that is on its side), so draw it that way, with the curve crossing the x-axis at x = 9 and the y-axis at y = -3 and y = +3. Then draw "x = y² - 9" as a right-opening parabola, pretty much the mirror of the first graph.

It's fairly obvious what the intersection points are, so you can see what the limits of integration will be. Since everything is "x in terms of y", and fairly nicely, why not just work it that way? That is, don't try to set up an integral of in terms of x; instead, integrate with respect to y:


. . . . .\(\displaystyle \large{\int_a^b{\,\mbox{(leftward-opening curve)}\,-\,\mbox{(rightward-opening curve)}\,}dy}\)


...where "a" and "b" are the negative and positive y-values of the intersection points.

Eliz.

 

[InterserveVB]

Thx

Thx. Works perfect.

 

[stapel]

Glad to hear it! :wink:

Eliz.