i) Find the volume of the solid under the graph of f(x; y) = 2+x+y that lies over the region in the xy-plane that is bounded by x2 + y = 1, x = 0 and y = 0.
ii) The region in i) is both vertically and horizontally simple. Write (but do not compute) the double integral in i) with the order of integration switched from the order you chose in i)
The volume under the surface `z=f(x;y) = 2 + x +y ` is given by the double integral
`int_(x in X)( int_(y in Y) f(x;y) dy ) dx ` ,
or equivalently (switching the order of integration)
`int_(y in Y) ( int_(x in X) f(x;y)dx )dy `
Now, `X ` is the valid domain for the variable `x ` , which here is `x>=0 ` *and* `|x|<=sqrt(1-y) ` . Similarly, `Y ` is the valid domain for `y `, which here is `y>=0 ` *and* `y <= 1 - x^2 ` . Since `x ` and `y ` are both positive, then the combined restriction is that neither can be greater than 1. This is important when calculating the outermost integral.
i) Suppose we choose to integrate over `y ` first, then we would write the volume `V ` such that
`V = int_0^1 ( int_0^(1-x^2) f(x;y) dy ) dx` `= int_0^1 ( int_0^(1-x^2) (2+x+y) dy)dx `
`= int_0^1 ( ((2+x)y + y^2/2)|_0^(1-x^2)) dx ` `= int_0^1((2+x)(1-x^2) + (1-x^2)^2/2) dx ` `= int_0^1 (1-x^2)(3/2+x-x^2/2) dx ` `= int_0^1 (3/2 + x - 2x^2 - x^3 + x^4/2) dx` `= (3/2x + x^2/2-(2x^3)/3 - x^4/4+x^5/10)|_0^1 `
`= 3/2 + 1/2 - 2/3 - 1/4 + 1/10 = (90+30-40-15+6)/60 = 71/60`
If we performed the integration the other way round we could calculate `V ` to be such that
`V = int_0^1 ( int_0^sqrt(1-y) f(x;y) dx) dy `
`= int_0^1 ( int_0^sqrt(1-y) (2+x+y)dx)dy ` `= int_0^1( (2+y)x + x^2/2)|_0^sqrt(1-y) dy `
`= int_0^1 ((2+y)sqrt(1-y) + (1-y)/2) dy `
As this integral would need to be done by parts, it is worth noting then that integrating with respect to `y ` first makes the calculation more straightforward in terms of algebra.