# Volume by double or triple integral?

I was trying to integrate the volume of a body blocked by $z=0,\; z=2x,\; x+y = 3$ and $y=0$ using the double integral... however it didn't work yet.

I'm convinced its a double integral and not a triple one. Any suggestion on how to do this?

• You do need a triple integral. Commented May 24, 2015 at 12:35
• find the area of crosssection on the surface $z=z_0$ first. Commented May 24, 2015 at 12:45
• Once a volume has been setup as a triple integral (the main difficulty of which is getting the nested limits of integration), performing the inner integration converts it to a double integration. So your intuition is right, but sometimes it is more expeditious just to use an approach you know will work and let the mechanical process create a better insight. Commented May 24, 2015 at 12:48

From the given equations of the blocking planes we can see that for given $x$, $y$ gets all values from $0$ to $3-x$ and $z$ from $0$ to $2x$. Values of $x$ are restricted to the interval from $0$ to $3$ so the volume is given by the triple integral $$V = \int_{0}^{3}\int_{0}^{3-x}\int_{0}^{2x} 1dzdydx = \int_{0}^{3}\int_{0}^{3-x}2x dydx = \int_{0}^{3} 2x(3-x) dydx = \int_{0}^{3} 6x-2x^2 dydx = 3x^2-\frac{2}{3}x^3|_{0}^{3} = 9.$$
The volume under a function (in our case $z=2x$) is a double integral: $$\int_0^3\int_0^{3-y}zdxdy=\int_0^3\int_0^{3-y}2xdxdy=\int_0^3\left[x^2\right]_0^{3-y}dy=\int_0^3\left(3-y\right)^2dy=-\frac{1}{3}\left[\left(3-y\right)^2\right]_0^{3}=-\frac{1}{3}\left[0-9\right]=3$$