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Find the volume $V$ of the described solid $S$.

The base of $S$ is the triangular region with vertices $(0, 0), (4,0)$, and $(0, 4)$.Cross-sections perpendicular to the x−axis are squares.

My answer is $V = 2π\left(\frac{27}{2} - \frac{27}{5}\right) = \left(\frac{π}{5} \right)(135 - 54) = \frac{81π}{5} $

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  • $\begingroup$ Your values are way off! $\endgroup$ – BenCarson2016 Nov 2 '15 at 0:58
  • $\begingroup$ Where did that $\pi$ come from? $\endgroup$ – Michael Biro Nov 2 '15 at 1:08
  • $\begingroup$ I think I have the wrong formula. Can you help me? $\endgroup$ – Cetshwayo Nov 2 '15 at 1:12
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According to your description of the solid, the base of the square in the plane perpendicular to the $x$-axis goes from the $x$-axis to the line $y=4-x$, so the side length is $4-x$. The area of that square is, of course, $y^2$, and the values of $x$ for those perpendicular planes run from $0$ to $4$.

Therefore the volume you want is

$$\int_0^4 (4-x)^2\,dx$$

You can finish from here.

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