# 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.

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I'm sorry, I wrote the question a bit wrong. I've updated it. –  user1405177 Oct 27 '12 at 16:39

$\int_0^1\int_{x^4}^x{(x+2y)\mathrm{d}y\mathrm{d}x}$ You have set correctly the integral but you did not integrate correctly.

Be careful in $\int_{x^4}^x{(x+2y)}\mathrm{d}y$ you integrate in respect of $y$ so you should find $$\int_{x^4}^x{(x+2y)}\mathrm{d}y=\left | xy+y^2 \right |_{x^4}^{x}=x^2+x^2-(x^5+x^{8})$$ And finally you get $\int_0^1 2x^2-x^5-x^{8}\mathrm{d}y=\frac{2}{3}-\frac{1}{6}-\frac{1}{9}=\frac{7}{18}$

You could verify your results here

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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*}

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