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Paladinen [302]
3 years ago
7

Evaluate the double integral. . ∫∫ y sqrt(x^2-y^2) dA, R={(x,y)|0≤y≤x, 0≤x≤1}. R. . Please explain

Mathematics
1 answer:
polet [3.4K]3 years ago
4 0
First we will evaluate: ( substitution: u = x² - y²,  du = - 2 y dy )
\int\limits^x_0 {y \sqrt{ x^{2} - y^{2} } } \, dy= \\   \frac{-1}{2} \int\limits^x_0 { u^{1/2} } \, du  =
=\frac{-1}{3} \sqrt{ (x^{2} - y^{2} ) ^{3} } ( than plug in x and 0 )
=- \frac{1}{3} (  \sqrt{( x^{2} - x^{2}) ^{3}  }  -  \sqrt{ (x^{2} -0 ^{2} ) ^{3} } =
= 1/3 x³ ( then another integration )
1/3\int\limits^1_0 { x^{3} } \, dx = 1/3 (  x^{4}/4)}= 1/3 ( 1 ^{4}/4 - 0^{4} /4 )
= 1/3 * 1/4 = 1/12
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