Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

= $\frac{8}{3}$

share|cite|improve this question
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 $0$. – Macavity Nov 16 '13 at 20:43
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. – André Nicolas Nov 16 '13 at 20:47
I think I know how to do it now but somehow I can't imagine what the picture of this problem looks like.. – IndyZa Nov 16 '13 at 20:59
up vote 1 down vote accepted

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.

share|cite|improve this answer

Your Answer


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.