Double Integral: Design Domain

[nycmath]

Design Domain and solve it.

[Eigenvalue]

Find the bounds, or limits of integration by setting the two functions equal
(x-1)²=1-x²
x²-2x+1=1-x²
2x²-2x=0
2x(x-1)=0
x=0, x=1

Rewrite the integral:
[tex]\int\limits_{0}^{1}[/tex] [tex]\int\limits_{(x-1)²}^{1-x²}[/tex](y)dy
[tex]\int\limits_{(x-1)²}^{1-x²}[/tex](y)dy

Evaluate the inner integral:
[tex]\frac{y²}{2} ]_{(x-1)²} ^{1-x²}[/tex]

=[tex]\frac{1}{2}[/tex](1-x²)²-[tex]\frac{1}{2}[/tex](x-1)⁴

=[tex]\frac{1}{2}[/tex]((1-x²)²-(x-1)⁴)

=[tex]\frac{1}{2}[/tex](1-2x²+x⁴-x⁴+4x³-6x²+4x-1)

=[tex]\frac{1}{2}[/tex](4x³-8x²+4x)

=2x³-4x²+2

Evaluate the outer integral:
[tex]\int\limits_{0}^{1}[/tex](2x³-4x²+2x)dx
=([tex]\frac{1}{2}[/tex]x⁴-[tex]\frac{4}{3}[/tex]x³+x²)[tex]]_{0 } ^{1 }[/tex]
Note: the lower bound for the outer integral is negligible, in this situation, as it evaluates to 0
[tex]\frac{1}{2}[/tex](1)⁴-[tex]\frac{4}{3}[/tex](1)³+(1)²

=[tex]\frac{1}{2}[/tex]-[tex]\frac{4}{3}[/tex]+1

=[tex]\frac{3-8+6}{6}[/tex]

=[tex]\frac{1}{6}[/tex]