# Computing volume by cross section method.

The base of the solid below is the region in the $xy$-plane bounded by the $x$-axis,the graph of y = $\sqrt{x}$ and the line $x = 3$. Find the volume of the solid.

Each cross-section of S perpendicular to the x-axis is a square with one side in the$xy$-plane.

• The limits of integration of all your slices will be the value of $\ x \$ where $\ \sqrt{x} \$ meets the $\ x-$ axis and $\ x \ = \ 3 \$ . The infinitesimal volume of each slice is the area of a square of side $\ \sqrt{x} \$ and a thickness $\ dx \$ . – colormegone Jan 19 '14 at 0:20
• The solid below? What do you mean? – Cameron Buie Jan 19 '14 at 0:26
• @RecklessReckoner Actually i don't understand this section from the professor at all. So first lets deal with the limits of integration. You said it will be the value of $x$ where $\sqrt{x}$ meets with $x$-axis and $x = 3$. So i set $\sqrt{x}$ equal to $3$ and then find $x$ ? – Out Of Bounds Jan 19 '14 at 0:27
• @CameronBuie That's what's written in the book. Actually the question is 4 parts and it said "The base of each solid below" but i just wrote the first part which is the square. – Out Of Bounds Jan 19 '14 at 0:29

So here's the graph of the relevant portion of $f(x) = \sqrt x$:

And here is a pic of the resulting solid with the same portion shaded for comparison:

Imagine taking slices of this solid as you move along the x-axis from 0 to 3. At each point x, the cross-sectional area $A(x)$ is $(\sqrt x)^2$. Now integrate $A(x)$ over the interval $[0, 3]$, so $V = \int _0^3(\sqrt x)^2dx = \int_0^3 xdx = \frac 9 2$

• Do i have to draw the graph ? In my case, here i don't know what $y = \sqrt{x}$ looks like. – Out Of Bounds Jan 19 '14 at 0:56
• Why $A(x) = (\sqrt{x})^2$ ? I know that the area of the square is $s^2$ but why is the side equal to $\sqrt{x}$ ? – Out Of Bounds Jan 19 '14 at 1:01
• You should know what $y = \sqrt x$ looks like now because I drew the graph for you. However, if you did not know what it looks like before then yes, you should definitely draw it out to get a feel for it because that is a very basic function that you should be very familiar with. As far as do you have to draw the graph for your homework: I don't know. Probably not, unless the instructions say so. I just drew that for you to provide clarity and understanding. Hope it helped. If I answered your question can you accept my solution an answer? – bgfriend0 Jan 19 '14 at 1:03
• The problem is telling you that each cross-section is a SQUARE, and we know that one of the square's side's length is equal to the height of $f(x)$ at whatever particular point we are concerned with at that moment. If ONE of the square's sides is $f(x) = \sqrt x$ then ALL FOUR of the square's sides have that same length. Thus at any point $x$ each side has length $\sqrt x$ and the square's area is $(\sqrt x)^2$. – bgfriend0 Jan 19 '14 at 1:07
• This is the pertinent part of the problem: "a square with one side in the xy-plane." We know the length of that one side: at $x$ it is $\sqrt x$. Thus we know the area of that cross-section at that point. – bgfriend0 Jan 19 '14 at 1:10

Since you are given cross-sections perpendicular to the x-axis, your limits of integration and integrand will be in terms of x:

$V=\int_0^3 A(x) \;dx=\int_0^3(s(x))^2\;dx$ where, as noted in the first comment above, $s(x)=\sqrt{x}$.

• Why the limits of the integration are from 0 to 3 ? – Out Of Bounds Jan 19 '14 at 0:42
• @Tennisman These are the values of x in the region which forms the base of the solid. (If we were using cross-sections perpendicular to the y-axis, we would use y-values instead.) – user84413 Jan 19 '14 at 0:46
• Why $s(x) = \sqrt{x}$ ? – Out Of Bounds Jan 19 '14 at 1:01
• @Tennisman Since you're taking slices perpendicular to the x-axis, the bottom side of the square is a vertical line segment with lower edge on the x-axis and top edge on the curve $y=\sqrt{x}$. – user84413 Jan 19 '14 at 1:04
• Can you please tell me how to do the same question but the cross section will be perpendicular to the x-axis and is an equilateral triangle with one side in the xy-plane and other part where the cross section will be perpendicular to the y-axis and is a square with one side in the xy plane – Out Of Bounds Jan 19 '14 at 1:41