Hello,
How do I calculate the triple integral for the solid created by the triple integral of "x dV" with limit "D", where D is bounded by paraboloid z=x^2+y^2 and the plane z=1?
Any help is much appreciated. Thanks.
The easiest way to approach this is to use cylindrical coordinates.
Consider the cross section of the solid when $\displaystyle z=1$: We have the circle $\displaystyle x^2+y^2=1\implies r^2=1\implies r=1$, after our conversion to cylindrical coordinates.
Thus, we can make the claim that $\displaystyle 0\leqslant r\leqslant 1$, $\displaystyle 0\leqslant\theta\leqslant2\pi$, and $\displaystyle 1\leqslant z\leqslant r^2$.
Taking into consideration that $\displaystyle x=r\cos\theta$ in the cylindrical coordinate system, we see that $\displaystyle \iiint\limits_D x\,dV=\int_0^{2\pi}\int_0^1\int_0^{r^2}\left(r\cos \theta\right) r\,dz\,dr\,d\theta=\int_0^{2\pi}\int_0^1\int_0^{r^ 2}r^2\cos\theta\,dz\,dr\,d\theta$
Can you continue on with the calculation?