Solve $\iint_D\sqrt{9-x^2-y^2}$ Where $D$ is the positive side of a circle of radius 3

Question

Solve $\displaystyle\iint_D\sqrt{9-x^2-y^2}$ Where $D$ is the positive side of a circle of radius 3 ($x^2+y^2=9,x\ge0,y\ge0$)

I tried to subsitute variables to $r$ & $\theta$:

$$x = r\cos\theta$$ $$y = r\sin\theta$$ $$E = \{0\le r\le3,0\le\theta\le\pi\}$$ $$\displaystyle\iint_D\sqrt{9-x^2-y^2} = \displaystyle\iint_EJ\sqrt{9-(r\cos\theta)^2-(r\sin\theta)^2}=\displaystyle\iint_Er\sqrt{9-(r\cos\theta)^2-(r\sin\theta)^2}$$

But have no clue on how to solve this new integral.

Answer 1

Be careful that the transformed domain $E$ should be defined by $0\le r\le3$ and $0\le\theta\le\pi/2$.

Now the integrand function is $$ r\sqrt{9-r^2(\cos^2\theta+\sin^2\theta)}=r\sqrt{9-r^2} $$ and this independent of $\theta$. For the integral over $dr$ use the substitution $r=3\sin u$.

Answer 2

It just is the half of the capacity of the ball: ($x^2+y^2+z^2=9)$.