The curves are $y=x^2, y=x$ and the cross sections are squares perpendicular to the $x-$axis such that the base of the squares is on the $xy-$plane.

My solution:

The area of the squares is given by $A=L^2$, where $L$ is the side of the square. We can find $L$ in function of $x$ computing the distance between the line and the parabola. The distance is $L=x-x^2$. Then, $A(x) = (x-x^2)^2.$

Hence, the desired volume is

$$V=\int_0^1 A(x) \mathrm{d}x = \int_0^1 (x-x^2)^2 \mathrm{d}x = \frac{1}{30}.$$

Is this correct? Is there another way to solve this kind of exercises?

  • 1
    $\begingroup$ Alternatively $$V=\int_0^1 (\sqrt x-x)^2 \mathrm{d}x = \frac{1}{30}$$ $\endgroup$
    – Quanto
    May 14 at 2:18


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.