Find the volume of the solid under the plane $x + 2y - z = 0$ and above the region bounded by $y=x$ and $y = x^4$. Find the volume of the solid under the plane $x + 2y - z = 0$ and above the region bounded by $y = x$ and $y = x^4$.
$$
\int_0^1\int_{x^4}^x{x+2ydydx}\\
\int_0^1{x^2-x^8dx}\\
\frac{1}{3}-\frac{1}{9} = \frac{2}{9}
$$
Did I make a misstep? The answer book says I am incorrect.

I think it should be calculated as
\begin{eqnarray*}
V&=&\int_0^1\int_{x^4}^x\int_0^{x+2y}dzdydx\\
&=&\int_0^1\int_{x^4}^x(x+2y)dydx\\
&=&\int_0^1\left.\left(xy+y^2\right)\right|_{x^4}^xdx\\
&=&\int_0^1(2x^2-x^5-x^8)dx\\
&=&\left.(\frac{2}{3}x^3-\frac{1}{6}x^6-\frac{1}{9}x^9)\right|_0^1\\
&=&\frac{7}{18}
\end{eqnarray*}
