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

integrate with respect to \(y\) first...let \(u=x^2-y^2\)

Answer 2

Do the chang of variable suggested in the inner iterval, then
$ \int_0^1 \int_0^x y \sqrt[x^2-y^2]dy dx=
\int_0^1 \frac {x^3} 3 dx = \left[ \frac {x^4}{12}]_0^1 = \frac 1 {12}$

Answer 3

\[\int\limits_0^1 \int\limits_0^x y \sqrt{x^2-y^2}dy dx= \int\limits_0^1 \frac {x^3} 3 dx= \left[ \frac {x^4}{12} \right]_0^1=\frac 1{12}\]

Answer 4

how did you get x^3/3? I'm unsure about the u-sub

Answer 5

eliassab, i have polar integration problem

Answer 6

Put \[u= x^2 -y^2\], then \[ du = -2 ydy\]
\[\int y \sqrt{x^2-y^2} dy =-\frac 1 2 \int \sqrt u du=-\frac{u^{3/2}}{3}\]
\[\left[ -\frac{\sqrt{x^2-y^2}^{3/2}}{3}\right]_0^x =\frac{x^3}3 \]

Answer 7

@perl what is it?

Answer 8

Thanks so much I understood!

Answer 9

The last line there is no Sqrt sign.
\[ \left[ -\frac{{x^2-y^2}^{3/2} }3\right] _{y=0}^{y=x}=\frac {x^3}3\] =

Answer 10

oh okay thanks