anonymous
  • anonymous
Calculus 1 AP Calculus BC guys please help me I give medals The base of a solid in the xy-plane is the circle x^2 + y^2 = 16. Cross sections of the solid perpendicular to the y-axis are squares. What is the volume, in cubic units, of the solid?
Mathematics
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
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anonymous
  • anonymous
@jim_thompson5910
anonymous
  • anonymous
@jigglypuff314
phi
  • phi
It's hard (for me) to visualize but the solid has slices that look like squares |dw:1450314441914:dw|

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anonymous
  • anonymous
Yeah, I can visualize it. but I don't know how to find the side of the rectangle
anonymous
  • anonymous
let me show you what I know
anonymous
  • anonymous
you now to find this volume of pics you need to find the limits of the integral? I know the integrals will go from -4 to 4 right?
anonymous
  • anonymous
and the formula of the volume are area * thickness. but I don't know how to derive the formula of the squares with only one graph. If this question had to graphs, I could just subtract x2 -x1 and find the side of the square
phi
  • phi
There is lots of symmetry. |dw:1450314718569:dw|
phi
  • phi
I would use x in the first quadrant \[ x= \sqrt{16-y^2} \] that is 1/2 of the side. i.e. the bottom is 2x the height is 2x and the area is 4x^2 or in this case \[ 4(16-y^2) \] I would integrate from y=0 to 4 and then double the answer for the full volume \[ 2\cdot 4\int_0^4 16-y^2 \ dy \]
anonymous
  • anonymous
oh okay thank you.
jim_thompson5910
  • jim_thompson5910
here is one example to help visualize the volume http://mathdemos.org/mathdemos/sectionmethod/gallerysqr.gif source: http://mathdemos.org/mathdemos/sectionmethod/sectiongallery.html ----------------------------------------------- here is another example (the left figure is probably the most similar) http://bowmanimal180.files.wordpress.com/2013/04/wpid-photo-apr-11-2013-718-am.jpg?w=604 source: http://bowmanwingspanson.com/2013/04/13/volume-in-calculus-conceptualizing-before-formalizing/
anonymous
  • anonymous
Thank you @jim_thompson5910 I understood it better now. Could you check if I did something right?
anonymous
  • anonymous
The base of a solid in the xy-plane is the circle x2 + y2 = 16. Cross sections of the solid perpendicular to the y-axis are squares. What is the volume, in cubic units, of the solid?
anonymous
  • anonymous
|dw:1450317183812:dw|
jim_thompson5910
  • jim_thompson5910
let me check
anonymous
  • anonymous
I know I have to add one to the function because it is collocated at x+1, but I am not sure if I should add to to the e or to the 1
anonymous
  • anonymous
I'm sorry wrong problem,
anonymous
  • anonymous
Which of the following integrals will find the volume of the solid that is formed when the region bounded by the graphs of y = e2x, x = 1, and y = 1 is revolved around the line y = −1.
jim_thompson5910
  • jim_thompson5910
oh I see, ok one sec
jim_thompson5910
  • jim_thompson5910
I don't agree with what you got
anonymous
  • anonymous
why?
jim_thompson5910
  • jim_thompson5910
let me get the drawing set up
anonymous
  • anonymous
ok
jim_thompson5910
  • jim_thompson5910
do you agree that we're spinning the green region around the red line y = -1 ?
anonymous
  • anonymous
yes
jim_thompson5910
  • jim_thompson5910
focus on just the line y = -1 and the curve e^(2x) what is the vertical line distance for any x on the interval 0 <= x <= 1
anonymous
  • anonymous
7.38
jim_thompson5910
  • jim_thompson5910
I'm looking for something in terms of x
jim_thompson5910
  • jim_thompson5910
pick any x in the interval 0 <= x <= 1 what is the vertical line distance, in terms of x, from y = -1 to y = e^(2x) ?
anonymous
  • anonymous
can I do an integral?
jim_thompson5910
  • jim_thompson5910
well this is helping to set up that integral
anonymous
  • anonymous
1+e^2x
jim_thompson5910
  • jim_thompson5910
yes correct since e^(2x) - (-1) = e^(2x) + 1
jim_thompson5910
  • jim_thompson5910
that's the radius of any circular cross section of the outer solid
jim_thompson5910
  • jim_thompson5910
the inner radius is going to be 1 - (-1) = 1+1 = 2 this is the distance from y = -1 to y = 1
jim_thompson5910
  • jim_thompson5910
Now use \[\Large \pi*\int\left[\left(\text{Outer radius}\right)^2-\left(\text{Inner radius}\right)^2\right]dx\]
anonymous
  • anonymous
ok, so to set up the integral now we can subtract the (e^2x +2)^2 - 1^2
anonymous
  • anonymous
from 0 to 1 right?
jim_thompson5910
  • jim_thompson5910
\[\Large \pi*\int\left[\left(\text{Outer radius}\right)^2-\left(\text{Inner radius}\right)^2\right]dx\] \[\Large \pi*\int_{0}^{1}\left[\left(e^{2x}+1\right)^2-\left(2\right)^2\right]dx\]
anonymous
  • anonymous
okay, thank you so much. I forgot to solve for the inner, but now I get how to solve for it.
jim_thompson5910
  • jim_thompson5910
no problem

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