determine the area enclosed by the given boundaries: y=10-x^2and y=x^2+2.
2 answers
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The two parabolas should be easy to graph to see what you are finding.
You will first need to find their intersection,
x^2 + 2 = 10 - x^2
2x^2 = 8
x = ± 2
The height of the region is (10-x^2) - (x^2+2)
= 8 - 2x^2
Because of the symmetry we will find the area from 0 to 2 and then double it.
Area = 2[integral] ( 8-2x^2) dx from 0 to 2
= 2 (8x - 2x^3/3)| 0 to 2
= 2[16 - 16/3 - 0]
= 64/3
You will first need to find their intersection,
x^2 + 2 = 10 - x^2
2x^2 = 8
x = ± 2
The height of the region is (10-x^2) - (x^2+2)
= 8 - 2x^2
Because of the symmetry we will find the area from 0 to 2 and then double it.
Area = 2[integral] ( 8-2x^2) dx from 0 to 2
= 2 (8x - 2x^3/3)| 0 to 2
= 2[16 - 16/3 - 0]
= 64/3
1 answer