Question

Please solve this qn: Evaluate the triple integral x.dx.dy.dz over the volume bounded by the paraboloid x= 4y^2 + 4z^2 and the plane x=4

Answer 1

hint: let r^2=y^2+z^2

Answer 2

Kinda hard to see clearly, but here's a rough sketch of the region you're concerned with (the space withing the paraboloid):

If we call this space \(D\), then
\[D:=\bigg\{(x,y,z)~:~0\le x\le4,~-\sqrt{\frac{x}{4}-z^2}\le y\le\sqrt{\frac{x}{4}-z^2},\\
~~~~~~~~~~~~~-\sqrt{\frac{x}{4}-y^2}\le z\le \sqrt{\frac{x}{4}-y^2}\bigg\}\]
I'd suggest a variation of converting to cylindrical, where \(dV=dx~dy~dz=r~dr~d\theta~dx\).

In cylindrical coordinates, you have
\[D:=\bigg\{(r,\theta,x)~:~0\le r\le1,~0\le\theta\le2\pi,~4r^2\le x\le4\bigg\}\]
\[\int\int\int_Dx~dV=\int_0^{2\pi}\int_0^1\int_{4r^2}^4xr~dx~dr~d\theta\]