Finding bounds for triple integral Let $S$ be the solid bounded by $$4x^2 +y^2 = 4, z+x=2, z=0$$, find the bounds for the triple integral of any function $f$ in the order $dzdydx$. The answers were that the bounds were $$0 \leq z \leq 2-x, -2\sqrt{1-x^2} \leq x \leq 2\sqrt{1-x^2}, -1 \leq y \leq 1 .$$
What I don't get is, suppose I couldn't draw the diagram (which I couldn't), how would I have known this? I thought we would have to solve the first and second equations simultaneously for an equation in terms of z and y, and then find bounds from there... but apparently not. How come?
 A: It's kind of hard to visualize I agree. Let's take the 3 equations separately and I assume that the $x$-$y$ plan is horizontal and the $z$ axis is vertical in space :


*

*$4x^2 + y^2 = 4$. We know this is an ellipse that has $x$-radius" $1$ and $y$-radius $2$. Since there is no constraint on $z$. It ranges over the whole space. So now in our mind we should have the idea of a long elliptical cylinder that is vertically unbounded.

*$z=0$. Clearly this is a plan and it is simply the $x$-$y$ plan. So this gives us a flat bottom to our cylinder.

*$x+z=2$. This will be the top of our cylinder. Unfortunately it is not flat. However it doesn't depend on $y$. So this means that if we fix some $y$ in our cylinder, the value will be constant there. Hence the cork of the cylinder will be a tilted plan "parallel" to the $y$-axis


Now based on that we need to chose the "free" variable, $x,y$ or $z$, that will range between fix number. Clearly $z$ is a bad choice since the cork is tilted and depends on $x$. So lets choose $x$ which ranges from $-1$ to $1$ according to the ellipse.
Then based on $x$ we can parameterize the cylinder as the equation suggests. (Solve $4x^2 + y^2 = 4$ for $y$. Note that you might have a typo in your question.)
Finally we need make $z$ ranges. Clearly it starts at the bottom of the cylinder at $0$ and then depending on $x$ goes to the tilted top. (Solve $x+z=2$ for $z$)
