What is the volume under the surface $z = f(x,y) = x^4 + xy + y^3$ over the rectangle $R = [1,2] \times [0,2]$.
I solved the double integral by integrating with respect to $x$ and then with respect to $y$. The answer I got was $67/5$ or $13.4$ but I am not sure whether or not I integrated incorrectly or in the incorrect order. Any feedback would be greatly appreciated.
On
Hint:
The volume is given by
$$ \int_1^2 dx \int_0^2 dy \int_0^{x^4+xy+y^3} dz $$
$$\begin{eqnarray} \int_1^2 dx \int_0^2 dy \int_0^{x^4+xy+y^3} dz &=& \int_1^2 dx \int_0^2 dy (x^4+xy+y^3)\\&=&\int_1^2 dx (2x^4+2x+4)\\&=& \Big[ \tfrac{2}{5} x^5 + x^2 + 4x\Big]_1^2 = \frac{97}{5} \end{eqnarray}$$
The integral that gives the volume is $$\iint z \, dA = \int_1^2 \int_0^2 x^4 +xy +y^3 \, dy \,dx.$$ Writing it as $$V = I_1 + I_2 + I_3$$ with $$\begin{align} I_1 & = \int_1^2 \int_0^2 x^4 \, dy \,dx, \\ I_2 & = \int_1^2 \int_0^2 xy \, dy \, dx, \\ I_3 & = \int_1^2 \int_0^2 y^3 \, dy \, dx, \end{align}$$ we find $I_1 = 62/5$, $I_2=3$ and $I_3 = 4$, therefore $$V = \frac{62}{5} +7 = \frac{97}{5}.$$