Evaluate the iterated integral $x= 0$ to $x=1$, $y= -\sqrt{1-x^2}$ to $y=\sqrt{1-x^2}$ and $z= 0$ to $z=2-x^2-y^2$ and $f(x,y,z)= \sqrt{x^2+y^2}$
What I know:
I know that I have to use cylindrical coordinates to make the integration easier to solve.
$$f(x,y,z)= r$$
$z= 0$ to $z=2-r^2$
$r= 0$ to $r=2$
$\theta= 0$ to $\theta=\pi $
I keep getting a negative number when the answer should be 7pi/15. Can anyone help me?
On
Where shall I begin - your region $R$ is simply half the intersection of cylinder $x^2+y^2=1$ and ratational paraboloid $z=2-x^2-y^2$ in the upper half plane $(z\geq 0)$. Cylindrical coordinates $(r,\varphi,z)$ are connected connected with cartesian with Jacobian $\left|\frac{\mathrm{d}(x,y,z)}{\mathrm{d}(r,\varphi,z)}\right|=r$, then
$$\iiint_R f(x,y,z)\,\mathrm{d}V=\int_{-\pi/2}^{\pi/2}\int_{0}^{1}\int_{0}^{2-r^2}r^2\,\mathrm{d}z\,\mathrm{d}r\,\mathrm{d}\varphi = \pi\int_{0}^{1}(2-r^2)r^2\,\mathrm{d}r=\pi\left(\frac{2}{3}-\frac{1}{5}\right)=\frac{7\pi}{15}$$
$$I=\int_{0}^{\frac{\pi}{2}}\int_{0}^{1}\int_{0}^{2-r^2}r^2\,dz\,dr\,d\theta+\int_{\frac{3\pi}{2}}^{2\pi}\int_{0}^{1}\int_{0}^{2-r^2}r^2\,dz\,dr\,d\theta$$ $$I=\pi\int_{0}^{1}(2r^2-4r^2)=\pi\left(\frac{2}{3}-\frac 15\right)=\frac{7\pi}{15}$$