Calculus III (or IV): Sketching Solids We are given:
Sketch the solid whose volume is given by the iterated
integral.
$$\int_{0}^{1}\int_{0}^{1}(4-x-2y)dxdy$$
I arrived at $-11/2$. However, we are asked to sketch the solid. I do not know how to do this freehand (or with TI-89 for what its worth). If someone could provide a reference to how it would help tremendously. 
 A: When Dealing with functions of one variable, for example, f(x), you have f(x)=stuff (stuff could be a polynomial in x, trig function in x, etc). Often, you are told that y=stuff, instead of f(x)=stuff, but really y=f(x). In functions of two variables, f(x, y), you can let f(x, y)=stuff, but this stuff is different from the others since stuff can be, for example, 4-x-2y. Its easy for some to let f(x, y) just equal another letter, z for example. Except now you have three orthogonal axis, (x axis, y axis, and f(x, y) axis). To sketch a solid, what I was taught was to let each z=0, or simply 0=4-x-2y, and sketch that in the xy-plane, then let x=0 and sketch z=4-2y in the zy-plane, and then finally letting y=0 and sketching z=4-x in the xz-plane. Each individual curve sketch you draw is called a trace, and thats how you sketch a solid in Euclidean 3-Space. Your drawing should look like a triangle since it is a flat surface.
A: If you have heard contour map, which may appear in your geography class. Think the solid as a mountain and draw the $\color{red}{contour\ line}$ first.
Steps to draw contour line(in a two dimension axes): 
$$z=f(x,y)=4-x-2y$$  let $z=0$ then draw $$0=4-x-2y$$ let$z=0.5$ then
draw $$0.5=4-x-2y$$ let$z=1$ then draw $$1=4-x-2y$$
Once you get a contour line graph then you can imagine to draw a solid, which seems not so hard to draw in this case.
Finally, if you want to check your graph, go to http://www.wolframalpha.com/input/?i=z%3D4-x-2y
A: You alaredy have the answer of how to sketch the graph, so I will help with evaluete your integral.

$$\int_{0}^{1}\int_{0}^{1}(4-x-2y)dxdy$$

$$=\int_{0}^{1}(4-\frac{1}{2}-2y)dy$$
$$=4y-\frac{y}{2}-\frac{2y^2}{2}\bigg|_{0}^{1}$$
$$=\frac52=\boxed{2.5}$$
