1)Find the volume of the solid over the rectangle $0\leq x\leq 1, 0\leq y\leq 3$ and bounded by the xy-plane and the plane z=x+y.
2)Suppose that a house is situated in one corner of a rectangular plot of land 300 feet by 200 feet. The contour of this plot of land is given by the equation $E(x,y)=10^{-4}(x^2-\frac{xy}{2})$ where the house is situated at the origin and the x-axis is the 300 foot length. It is desired to level this property at the level of the house. How much fill has to be removed (or brought in) to accomplish this?
Solution:- Both the question can be solved using double integration with respect to x and y.
Answer to first question is 6 and answer to second question is 135000 cubic feet.
Were we to ask for the average elevation property above house level, we divide this by the area of the property, getting $\frac{135000}{60000}=2.25$ feet. I didn't understand what does this mean?
Secondly, though i got the answers to both the questions, i want to understand both answers to these questions pictorially.
So, any member may answer to these questions using graphics in 2-dimensional Cartesian space.
The L.H.S. figure depicts the graph of the answer to first question. The R.H.S. figure does not relate to these questions. So please ignore it.
Secondly, how to plot graph of the answer to the second question manually on a piece of graph paper?
A basic double integral of a function can be thought of as the signed volume between the function and the coordinate axes. Single integration is usually thought of as the signed area under a curve, this area can be thought of as a slice of a volume and the second integral can be thought of as summing up these slices which yields a volume. The rectangle you use as bounds on a double integral then works as a boundary to this volume.
The first question is pretty trivially answered by my explanation and a quick double integral verifies the answer: $$ \int_0^3 \int_0^1 x+y dxdy \\ \int_0^3 (\frac{1}{2}x^2+yx \bigg\rvert^1_0)dy \\ \int_0^3 \frac{1}{2} + ydy \\ \frac{1}{2}y+\frac{1}{2}y^2\bigg\rvert^3_0 \\ \frac{3}{2} + \frac{9}{2} = \frac{12}{2} = 6 $$ So when it comes to your problem the reason that double integration can solve it is that you want to find the volume of this contour plot that is above and below, the house level.