Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

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

2 Answers 2

up vote 0 down vote accepted

$\int_0^1\int_{x^4}^x{(x+2y)\mathrm{d}y\mathrm{d}x}$ 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

share|improve this answer

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*}

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.