# Volume of a body bounded by various functions

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?

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

I hope the folowing sketch clarifies your doubts. 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*}

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

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,

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