1
$\begingroup$

How would you calculate the volume of a body bounded by surfaces $z=\log y, z=0, y=e, x=2, x=4$?

I'm having trouble figuring out the limits of integration.

So, $x$ and $z$ are pretty obvious, but $y$ is not.

I was thinking that $y$ needs to go from $0$ to $e$ since the graph of the function $\log y$ gets arbitrarily close to zero.

So my integral would look like $$\int_2^4\int_0^e\int_2^{\log y}dzdydx$$

Is this correct?

$\endgroup$
  • $\begingroup$ The integral in the present form is negative: $\int_{2}^{4}\left( \int_{0}^{e}\left( \int_{2}^{\log y}dz\right) dy\right)dx= -4e$. $\endgroup$ – Américo Tavares Jun 10 '17 at 20:04
1
$\begingroup$

I hope the folowing sketch clarifies your doubts.

enter image description here

The body is limited below by $ z=0 $ and above by $ z=\log y $. On the right by the plane $ y=e $ as shown in the picture. So the limits of integration with respect to $ y $ are $ y=1 $ ( due to the condition $ 0\le z$), $ y=e $; and with respect to $ z $, $ z=0, z=\log y $. The limits with respect to $ x $ are correct.

ADDED. The computation is easy and yields a positive value, as it should be, and not $-4e$ as per your integral (see my comment). The volume is

\begin{equation*} V=\int_{2}^{4}\left( \int_{1}^{e}\left( \int_{0}^{\log y}dz\right) dy\right) dx=\int_{2}^{4}\left( \int_{1}^{e}\log y\;dy\right) dx=\int_{2}^{4}1\;dx= 2. \end{equation*}

$\endgroup$
  • $\begingroup$ That's very helpful, thank you very much. $\endgroup$ – windircurse Jun 11 '17 at 8:32
  • 1
    $\begingroup$ @windircurse Glad to help. $\endgroup$ – Américo Tavares Jun 11 '17 at 9:18
1
$\begingroup$

You saw how simple is the part for $x$: two parallel planes and all other surfaces perpendicular. But this reduces the problem to an usual one in dimension two, but in the plane $yz$. In this kind of problems the idea in to find the intersection of the several curves defining the surface.

$z=0$ and $z=\log y$, intersecting at $z=0;\;y=1$. The line $y=e$ cuts somewhere the curve $z=\log y$ ($z=1$, indeed), so, this is the upper limit for $y$. Then,

$$A=\int_2^4\int_1^e\int_0^{\log y}dzdydx$$

$\endgroup$

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.