Sketch the region enclosed by the curves x= 49-y^2 and x = y^2 - 49. Decide whether to integrate with respect to x or y. Then find the area of the region.

Answer 1

sketching should have been no problem
both have axes on the x-axis,
one has vertex (-49,0) and opens to the right
the other has vertex (49,0) and opens to the left
There is perfect symmetry and they meet at (0,7) and (0,-7)
I would take horizontal slices, thus integrate with respect to y
Also go from y = 0 to y = 7 and double that answer

area = 2(integral) (49 - y^2 - (y^2-49)) dy from 0 to 7
= 2(integral) (98 - 2y^2) dy from 0 to 7
= 2 [98y - (2/3)y^3] from 0 to 7
= 2(686 - (2/3)(343) - 0 )
= 2(1372/3)
= 2744/3 or 914 2/3 units^2

confirmation:
http://www.wolframalpha.com/input/?i=area+x%3D+49-y%5E2+%2C+x+%3D+y%5E2+-+49

I don't know why Wolfram turned the graph sideways, notice the y-axis is horizontal and the x-axis is vertical