Question

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

Answer 1

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