A solid in the first octant is bounded by $y=x^2, y=0, z=0$ and $x+y+z=6$. I am tasked to use triple integration to compute its volume using order $dz \thinspace dx \thinspace dy$.
My work. Consider the solid to be of type $xy$ bounded below by the plane $z=0$ and bounded above by $z=6-x-y$. Its projection $R$ onto the $xy$-plane is given by the region bounded by $y=x^2$ and $x+y=6$. If it is to be classified of type II (using horizontal strips), then we would have to cut the region at $y=4$. Thus, setting up the integral, we have
$$\int_0^4 \int_0^{\sqrt{y}} \int_0^{6-x-y} dz \ dx \ dy + \int_4^6 \int_0^{6-y} \int_0^{6-x-y} dz \ dx \ dy$$
which yields the answer $\frac{248}{15}$. Did I do something wrong here?
Best Answer
Please note the bound region is outside the parabolic cylinder as one of the bounds is plane $y = 0$. See the diagram of the projection in xy-plane. What you are considering is the region shaded on the left. But what I think the question intends is the region on the right. It is the region bound by parabolic cylinder $y = x^2$ and planes $x+y+z = 0, y = 0, z = 0$.
For every point in the region in xy-plane, $0 \leq z \leq 6 - x - y$
Now for the region in xy-plane, if we integrate wrt. $x$ first, we need one integral.
$\sqrt y \leq x \leq 6 - y$
For bound of $y$, we see intersection of $x + y = 6$ and $y = x^2$
$x^2 + x - 6 = 0 \implies (x+3)(x-2) = 0$
As we are in first octant, $x= 2, y = 4$.
So integral becomes,
$\displaystyle \int_0^4 \int_{\sqrt y}^{6-y} \int_0^{6-x-y} dz \ dx \ dy = \cfrac{292}{15}$