1
$\begingroup$

The problem is stated below:

Let $V$ be volume bounded by surface $y^2=4ax$ and the planes $x+z=a$ and $z=0$.

Express $V$ as a multiple integral, whose limits should be clearly stated. Hence calculate $V$.

Progress

I want to find out the limits of the multiple integral that's needed to calculate $V$.

I'm guessing that: $x=a-z$, $x=(y^2)/4a$

$$y= \pm \sqrt{4ax},\quad z=0,\quad z=a-x $$

but it seems like I've used the upper plane equation twice?

Also, it would really help if we could compare our answer volumes to check that this is right from the start!

Thanks everyone!!! :)

$\endgroup$
17
  • 1
    $\begingroup$ We don't just blindly compute homework problems here. What's your question? $\endgroup$
    – apnorton
    Dec 9, 2014 at 16:21
  • $\begingroup$ Some of us do "blindly" do HW problems. But it's a good idea, pikayenga, if you want folks to help out, to say "What I've done so far is XYZ, but now I don't know how to find the limits of integration", or "Should I be integrating with respect to $xy$, $xz$, or $zy$, or perhaps in some polar coordinate system? I don't know how to tell in advance what will work." Also: try to learn how to use LaTeX to write math nicely (I've done it for you on this first posting.) $\endgroup$ Dec 9, 2014 at 16:24
  • $\begingroup$ So: Why not edit this posting and tell us what you've done so far? $\endgroup$ Dec 9, 2014 at 16:24
  • $\begingroup$ @pikayenga is the problem finding the limits or turning the volume into an integral? $\endgroup$
    – MickG
    Dec 9, 2014 at 16:25
  • $\begingroup$ I gotta say finding those limits is not that easy at all :). $\endgroup$
    – MickG
    Dec 9, 2014 at 16:31

2 Answers 2

2
$\begingroup$

Sometimes a picture is worth a thousand words. Eventually, a math student should try plot graphs in their minds with out the use of software (although, I've always preferred Play Doh because after solving the problem I could always make little fishies and green froggies).

Geometrically, Your problem should look like this... enter image description here

We see that $x=a$ when the upper plane intersect the $x-y$ plane. We also see that when $x=t$ $$y^2=4at$$ $$y=\pm2\sqrt{at}$$

So the integral becomes $$\begin{array}{lll} \int^a_0\int_{-2\sqrt{ax}}^{2\sqrt{ax}}(a-x)dydx&=&\int^a_0(a-x)(4\sqrt{ax})dx\\ &=&\int^a_0(4a^\frac{3}{2}x^\frac{1}{2} - 4a^\frac{1}{2}x^\frac{3}{2})dx\\ &=&(\frac{2}{3}\cdot 4a^\frac{3}{2}a^\frac{3}{2} - \frac{2}{5}\cdot 4a^\frac{1}{2}a^\frac{5}{2})\\ &=&8a^3(\frac{1}{3}-\frac{1}{5})\\ &=&8a^3(\frac{2}{15})\\ &=&\frac{16}{15}a^3\\ \end{array}$$

$\endgroup$
6
  • $\begingroup$ Thank you so much! This is very useful! :) $\endgroup$ Dec 9, 2014 at 18:17
  • $\begingroup$ isn't the lower limit stated to be at z=0? How is it on the parabola? $\endgroup$ Dec 9, 2014 at 19:11
  • $\begingroup$ what did you get as the volume eventually? $\endgroup$ Dec 9, 2014 at 19:21
  • $\begingroup$ see my updated answer $\endgroup$
    – John Joy
    Dec 9, 2014 at 19:44
  • $\begingroup$ you're right, my mistake. The limits for $z$ do in fact go from $0$ to $a$. $\endgroup$
    – John Joy
    Dec 9, 2014 at 19:51
1
$\begingroup$

OK so first of all we must turn "bounded by" into inequalities. If you picture the three surfaces, you realize the only bounded region is:

$$\left\lbrace y^2\leq 4ax,z\geq0,x+z\leq a\right\rbrace=\left\lbrace y\in[-\sqrt{4ax},\sqrt{4ax}],x,z\geq0,x\leq a-z\right\rbrace.$$

The limits for $y$ are explicit and depend on $x$, so we put the $dy$ integral inside. With that, $x,z$ don't depend on $y$, so we will have one from 0 to $a$, and the other in such a way that the sum is less than $a$. I wrote the equation in terms of $x$ to suggest my approach, but swapping the integrals shouldn't give any change.

How do we see that is the region? Well we have two planes and a 'parabolc prism'. The region outside the prism will be unbounded, which for example gives $x\geq0$ and the $y$ inequality. Under the $z=0$ plane, the region can go to infinity in $x$, since there is no upper bound for it from any of the surfaces. So $z\geq0$. The other bit, $x\leq a-z$, is just the bounding of the plane, which has to be an upper bound since the other plane is a lower one and the 'prism' doesn't touch $z$ in any way.

Hope I've been clear.

$\endgroup$
6
  • $\begingroup$ Why does the code not get rendered correctly? I only see the right half of the sets on my mobile!! $\endgroup$
    – MickG
    Dec 9, 2014 at 17:16
  • 1
    $\begingroup$ Thank you so much for helping!!! @MickG Can I just clarify that you are suggesting the limits are: 0<x<(a-z), -sqrt(4ax)<y<+sqrt(4ax), 0<z<a ?? $\endgroup$ Dec 9, 2014 at 17:33
  • $\begingroup$ I guess I'd suggest weak inequalities though @pikayenga :). Shouldn't change much anyway. Do learn some LaTeX though :). $\endgroup$
    – MickG
    Dec 9, 2014 at 17:55
  • $\begingroup$ OK! I will try and learn to use LaTeX from now on! Thanks for your kind advice and help! :) $\endgroup$ Dec 9, 2014 at 18:12
  • $\begingroup$ This is a good start. In particular the sections about maths. Note: this site uses MathJax, which is part of LaTeX, not the whole of LaTeX, so don't be too surprised if some commands described in the linked wikibook do not work here :). MathJax code is always between dollars $…$ or double dollars $$…$$ (for "displayed" equations, which are centered on a new line. You're welcome. $\endgroup$
    – MickG
    Dec 9, 2014 at 20:23

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .