Finding volume using integrals.

  • Thread starter MarcL
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  • #1
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Huhh First of all I'm sorry if this is the wrong question ( didn't know if this was considered pre cal. I have a gut feeling it is :P)

Homework Statement



Let R be the region bounded by y=tan x, y=0 and x = Pi/4
Find the volume of the solid whose base is region R and whose cross-section is a square perpendicular to the x-axis

Homework Equations



Huh well I don't know whether its cylindrical shell:
ab Pi(R)(Height)dx
or disk method
ab A(x) dx


The Attempt at a Solution


I can't even figure out what the graphs look like. The wording is really confusing. I tried using a pyramid shape with a square base because it says the cross section is a square and using similar triangle ( I think its called like that...) but I couldn't get the equation for the formula.
 

Answers and Replies

  • #2
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6,857
Huhh First of all I'm sorry if this is the wrong question ( didn't know if this was considered pre cal. I have a gut feeling it is :P)
No, this is calculus, not precalculus.

Homework Statement



Let R be the region bounded by y=tan x, y=0 and x = Pi/4
Find the volume of the solid whose base is region R and whose cross-section is a square perpendicular to the x-axis

Homework Equations



Huh well I don't know whether its cylindrical shell:
ab Pi(R)(Height)dx
or disk method
ab A(x) dx


The Attempt at a Solution


I can't even figure out what the graphs look like.
First off, you need to figure out what region R looks like. Surely you have studied the trig functions before taking this class.
The wording is really confusing. I tried using a pyramid shape with a square base because it says the cross section is a square and using similar triangle ( I think its called like that...) but I couldn't get the equation for the formula.

The base isn't square. The vertical cross-sections are square.

BTW, why do you start some of your sentences with "Huh" or "Huhh"?
 
  • #3
170
2
I don't know, force of habit when speaking out load transferred to my typing skills... thankfully not in essays. Plus English isn't my first or second language. And yeah i did study trig function I know what it looks like. I assumed the base was squared because all the other problems I did, whenever I took the cross section from the shape, it was always the same shape as the base ( so lets say a sphere, was a circle, a square-based pyramid was a square, etc...)

Now for the graph It goes from 0 to pi/4 which is a curve on the first quadrant. However with my assumption from before, If i rotate that shape around the x-axis, I can only take a circle as the cross section as my shape looks something very similar to a cone, no?
 
  • #4
35,118
6,857
I don't know, force of habit when speaking out load transferred to my typing skills... thankfully not in essays. Plus English isn't my first or second language. And yeah i did study trig function I know what it looks like. I assumed the base was squared because all the other problems I did, whenever I took the cross section from the shape, it was always the same shape as the base ( so lets say a sphere, was a circle, a square-based pyramid was a square, etc...)

Now for the graph It goes from 0 to pi/4 which is a curve on the first quadrant. However with my assumption from before, If i rotate that shape around the x-axis, I can only take a circle as the cross section as my shape looks something very similar to a cone, no?
There is NO rotation going on!! The base of the shape is that graph of y = tan(x), 0 ≤ x ≤ ##\pi/4##.

To draw the solid, you need to be able to draw something in three dimensions.
1. Pick a point x in the interval.
2. Draw a line from the x-axis straight over to the curve. (I'm assuming that your graph is in the x-y plane in 3D.
3. Now draw a square whose vertical height is the same as its base. The square should be perpendicular to the x-axis.

Pick another point in the interval, and repeat the steps above.
Draw enough squares so that you have a good idea what the solid looks like.
 
  • #5
170
2
Ah! I see it now!! so my limits of integration would be from 0 to pi/4. and my Area would be tan^2(x)?
 
  • #6
35,118
6,857
Ah! I see it now!! so my limits of integration would be from 0 to pi/4. and my Area would be tan^2(x)?
If you mean this --
$$ \int_0^{\pi/4} \tan^2(x)~dx$$

then, yes.
 
  • #7
170
2
Thank you so much! and thanks for not giving me the answer. I like understanding what I'm doing! And that is what I meant!
 

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