# Triple integral bounds

1. Let $W$ be the region bounded by the planes $x = 0$, $y = 0$, $z = 0$, $x + y = 1$, and $z = x + y$.

2. $(x^2 + y^2 + z^2)\, \mathrm dx\, \mathrm dy\, \mathrm dz$; $W$ is the region bounded by $x + y + z = a$ (where $a > 0$), $x = 0$, $y = 0$, and $z = 0$.

$x,y,z$ being $0$ is throwing me off because I'm not sure how to graph it and get a bounded area; it seems like it would be infinite to me. What would the bounds for the triple integrals be?

• $x,y,z =0$ creates floors for you. These will be your lower bounds – Doug M Feb 2 '17 at 4:50

1) $$\int_{x=0}^{x=1}\int_{y=0}^{y=1-x}\int_{z=0}^{z=x+y}\ dzdydx$$

2) $$\int_0^{x=a}\int_0^{y=a-x}\int_0^{a-x-y} (x^2 + y^2 + z^2) \;dz\;dy\;dx$$

1) give you nothing to integrate. But we can still find the limits

$z = 0$ lower limit for $z$

$z = x+y$ upper limit for $z$

$y = 0$ lower limit for $y$

$y = 1-x$ upper limit for $y$

$x = 0$ lower limit for $x$

$x = 1$ upper limit for $x$

For the last one, because that is where $y = 0$ intersects $x+y = 1$

2)

$\int_0^a\int_0^{a-x}\int_0^{a-x-y} x^2 + y^2 + z^2 \;dz\;dy\;dx$

• why did you choose a to be 1? also why are upper bounds for y and x 1? – user3427042 Feb 2 '17 at 5:08