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Find the volume of the solid whose base is a triangular region with vertices (0, 0), (2, 0) and (0, 2) if the cross-sections perpendicular to the Y -axis are squares.

My problem is that I can't figure out how the shape will be like... I need some images.

What I think is

Volume = $\int\limits_a^b Area\,dy$

=$\int\limits_a^b base^2\,dy$

= $\int\limits_0^2 y^2\,dy$

= $\frac{8}{3}$

?????

[2nd Edit]

Should be like this?

Volume = $\int\limits_a^b Area\,dy$

=$\int\limits_a^b base^2\,dy$

= $\int\limits_0^2 x^2\,dy$ ; Since y = -x +2 or x = 2-y

= $\int\limits_0^2 (2-y)^2\,dy$

= $\int\limits_0^2 y^2-4y+4\,dy$

= $[\frac{y^3}{3} - 2y^2 +4y]$ from 0 to 2

= $\frac{8}{3}$

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  • $\begingroup$ In your integral, the formula for the base is incorrect. For e.g. when $y=0$, the base is $x=2$. Similarly, when $y=2$, the base should be $0$. $\endgroup$ – Macavity Nov 16 '13 at 20:43
  • $\begingroup$ The line joining $(2,0)$ and $(0,2)$ has equation $x+y=2$. The cross-section "at" $y$ is an $x\times x$ square. But $x=y-2$, so you want to integrate $(2-y)^2$. It so happens that this yields precisely the same numerical answer as your calculation. But your setup is not right. $\endgroup$ – André Nicolas Nov 16 '13 at 20:47
  • $\begingroup$ I think I know how to do it now but somehow I can't imagine what the picture of this problem looks like.. $\endgroup$ – IndyZa Nov 16 '13 at 20:59
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If the cross-sections perpendicular to the $Y$-axis are squares, then we know that the cross-sectional area at a height $y$, will be given by the square distance from the Y-axis to your line $y=2-x$. This distance is just $x$. If we integrate over the $y$-direction we can now find the volume of your object.

So, volume = $\int_{0}^{2} x^2 dy$. Here we have the mismatch of a variable in $x$ and our integration being over $y$, we so we the equation of the line to substitute $x = (2-y)$, and we find that the volume is $\int_{0}^{2} (2-y)^2 dy$.

I think you can take it from here.

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