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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.

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  • $\begingroup$ 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 \ $ . $\endgroup$ – colormegone Jan 19 '14 at 0:20
  • $\begingroup$ The solid below? What do you mean? $\endgroup$ – Cameron Buie Jan 19 '14 at 0:26
  • $\begingroup$ @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 $x$ ? $\endgroup$ – Out Of Bounds Jan 19 '14 at 0:27
  • $\begingroup$ @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. $\endgroup$ – Out Of Bounds Jan 19 '14 at 0:29
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So here's the graph of the relevant portion of $f(x) = \sqrt x$:

enter image description here

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

enter image description here

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 $V = \int _0^3(\sqrt x)^2dx = \int_0^3 xdx = \frac 9 2$

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  • $\begingroup$ Do i have to draw the graph ? In my case, here i don't know what $ y = \sqrt{x} $ looks like. $\endgroup$ – Out Of Bounds Jan 19 '14 at 0:56
  • $\begingroup$ 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} $ ? $\endgroup$ – Out Of Bounds Jan 19 '14 at 1:01
  • $\begingroup$ 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? $\endgroup$ – bgfriend0 Jan 19 '14 at 1:03
  • $\begingroup$ 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 $(\sqrt x)^2$. $\endgroup$ – bgfriend0 Jan 19 '14 at 1:07
  • $\begingroup$ 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. $\endgroup$ – bgfriend0 Jan 19 '14 at 1:10
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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, $s(x)=\sqrt{x}$.

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  • $\begingroup$ Why the limits of the integration are from 0 to 3 ? $\endgroup$ – Out Of Bounds Jan 19 '14 at 0:42
  • $\begingroup$ @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.) $\endgroup$ – user84413 Jan 19 '14 at 0:46
  • $\begingroup$ Why $ s(x) = \sqrt{x} $ ? $\endgroup$ – Out Of Bounds Jan 19 '14 at 1:01
  • $\begingroup$ @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 $y=\sqrt{x}$. $\endgroup$ – user84413 Jan 19 '14 at 1:04
  • $\begingroup$ 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 $\endgroup$ – Out Of Bounds Jan 19 '14 at 1:41

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