How do I find the bounds of integration for this?
I assume I need to find the standard equation for this so I put it like so:
$$\frac{x}{7}+\frac{y}{9}+\frac{z}{2}=1$$
and so I solve for $z$. Then I solve for $y$ by making $z=0$ And then repeat for X.
And so here are my bounds:
$$\int _0^7\int _0^{9-\frac{9x}{7}}\int _0^{-\frac{2x}{7}-\frac{2y}{9}}(z)\,dz\,dy\,dx$$
Which equals $\frac{63}{2}$ which isn't the correct answer.
How would I find my bounds of integration here?
Thank you.
The outermost integral can be with respect to $x$, as the cross-sections perpendicular to the $x$-axis are all the same. Its bounds are $0$ to $a$.
The middle integral will be taken with respect to $y$ and be from $0$ to $b$. But the range of the innermost integral (in $z$) depends on whatever value of $y$ is "in effect" when the innermost integral is evaluated: we have a constant lower bound of $0$, but the upper bound linearly varies from $0$ at $y=0$ to $c$ at $y=b$. The function describing this is $yc/b$. Thus, our final triple integral is $$\int_0^a\int_0^b\int_0^{yc/b}z\,dz\,dy\,dx$$