I have to integrate and describe of what region in $R^3$ they represent the volumes.
1) $\displaystyle\int_{0}^{2} \int_{-2}^{2} 4-x^2-y^2 dxdy$
2) $\displaystyle\int_{-2}^{3} \int_{0}^{1} |x|\sin(\pi y) dydx$
Solution:
1)$\int_{0}^{2} 4x-\frac{x^3}{3}-xy^2 dy$ ;integrating w.r.t x
$\int_{0}^{2} \frac{32}{3} - 4y^2 dy$ ;inserting valuyes of x
$\frac{32y}{3}-\frac{4y^3}{3}$ ; integrating w.r.t y
$\frac{32}{3}$ ; inserting values of y
The region should be a paraboloid over a rectangle $R$={${(x,y)\epsilon R^2 | 0<=y<=2,-2<=x<=2}$}.
Is this correct? because I drew the graph and found that the volume is not always non-negative. I have similarly found the volume for part 2, and its region should be something over a rectangle R$={${(x,y)\epsilon R^2 | 0<=y<=1,-2<=x<=3}$} but I don't know what that something is.