Evaluate the triple integral 
where E is the solid bounded by the cylinder 
and the planes
and 
in the first octant.
I tried really hard to get the write answer but I am still struggling in the picture so I could not find the limits
[Math] triple integral bounded by cylinder
integrationmultivariable-calculus
Related Solutions
You want it in the first octant so we have $$x\ge0, y\ge0,z\geq0$$ The lower limit of $y$ is zero but another limit of it is ruled by $y^2+z^2=1$. Since $z=0$ is one of our boundary , so $y^2=1$ and so $y=1$. Clearly, the latter equation gives us the restriction for $z$, that is: $$z\big|_0^{\sqrt{-y^2+1}}$$ Note that here again we have $z\ge 0$. For $x$ is the same, that is $$x\big|_0^{2-y}$$ since we already know that $x+y=2$.
Use spherical coordinates for the sphere in the first orthant:
$$\;0\le r\le1,\,0\le\theta\le\frac\pi2,\,0\le\phi\le\frac\pi2\;$$
Observe that here $\;\theta\;$ is the azimuth angle and $\;\phi\;$ is the altitude, or vertical, angle. Then the integral is
$$\int_0^1\int_0^{\pi/2}\int_0^{\pi/2}\overbrace{r\cos\theta\sin\phi}^{=x}\cdot \overbrace{r\sin\theta\sin\phi}^{=y}\cdot \overbrace{r\cos\phi}^{=z}\cdot \overbrace{r^2\sin\phi}^{=\text{Jacobian}}\, d\phi\,\mathrm d\theta\,\mathrm dr=$$
$$=\int_0^1\int_0^{\pi/2}\int_0^{\pi/2} r^5\cos\theta\sin\theta\cos\phi\sin^3\phi\, d\phi\,\mathrm d\theta\,\mathrm dr=$$
$$=\left.\frac16r^6\right|_0^1\cdot\left.\frac12\sin^2\theta\right|_0^{\pi/2}\cdot\left.\frac14\sin^4\phi\right|_0^{\pi/2}=\frac16\cdot\frac12\cdot\frac14=\frac1{48}$$
Best Answer
you can check if this is correct: $$\int_{\theta=0}^{\pi/2}\int_{r=0}^3\int_{x=0}^{\frac{rcos\theta}{3}}r^2sin\theta dxdrd\theta$$