CoconutJJ
  • CoconutJJ
The bounded area by the graph y=sqrt(x), y=x-4 and the x axis is rotated along the x axis to form a solid. Use Integration By Shells to find the Volume of the Solid.
Mathematics
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
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misssunshinexxoxo
  • misssunshinexxoxo
This might help utilize the "Shell" method; https://www.khanacademy.org/math/integral-calculus/solid_revolution_topic/solid_of_revolution/v/solid-of-revolution-part-5
CoconutJJ
  • CoconutJJ
|dw:1435377786774:dw| This is the bounded area
CoconutJJ
  • CoconutJJ
Wait so how does the integral change when you rotate it around the x axis instead of y?

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misssunshinexxoxo
  • misssunshinexxoxo
https://www.wyzant.com/resources/lessons/math/calculus/integration/finding_volume
misssunshinexxoxo
  • misssunshinexxoxo
With "washer method"
misssunshinexxoxo
  • misssunshinexxoxo
To rotate about the y-axis, swap the roles of x and y throughout the above discussion. To rotate about a different axis, change coordinates so that the axis of rotation is one of the coordinate axes, and then do the computation.
misssunshinexxoxo
  • misssunshinexxoxo
Follow rules of "shell" and "washer" method to guide throughout finding volume
anonymous
  • anonymous
been a while, and I was never great at these so take this with a grain of salt. I believe you have to change the formulas around and integrate in terms of y. |dw:1435378239112:dw|
anonymous
  • anonymous
and then the limits will be the y coordinates of the intersection points
CoconutJJ
  • CoconutJJ
so then the thickness of the cylinder is dy?
CoconutJJ
  • CoconutJJ
|dw:1435378522268:dw|
UsukiDoll
  • UsukiDoll
@peachpi it's been a while for me too.
misssunshinexxoxo
  • misssunshinexxoxo
Peachpi seems absolutely right; this is a tough area
anonymous
  • anonymous
this is a solid. so there's no hole, right?
anonymous
  • anonymous
the area lies on the axis it's being rotated on
UsukiDoll
  • UsukiDoll
but eventually we have to integrate this thing... who is going to be a and who is going to be b on the integral symbol ?
CoconutJJ
  • CoconutJJ
well they intersect when x=6.562
anonymous
  • anonymous
yeah, but the integral is in y. a is 0. b you solve y² = y + 4 to get
CoconutJJ
  • CoconutJJ
How is the integral with respect to y ?
UsukiDoll
  • UsukiDoll
maybe it has something to do with the x axis being rotated so we can't use that? I have no idea x.x
UsukiDoll
  • UsukiDoll
oh I see where the a and b for the integral is |dw:1435378968903:dw|
UsukiDoll
  • UsukiDoll
those intersection points are the values for a and b
CoconutJJ
  • CoconutJJ
|dw:1435378996664:dw|
UsukiDoll
  • UsukiDoll
well a = 0 is correct.. but for the b we need to find the value of that intersection point but as @peachpi mentioned we have to solve " y² = y + 4 to get"
anonymous
  • anonymous
I think it is simpler to do it horizontally than vertically, which would require two integrals.
UsukiDoll
  • UsukiDoll
hmmm y^2-y-4 = 0 and then factor ? OH NO! I DO NOT WANT TO DO 2 INTEGRALS
UsukiDoll
  • UsukiDoll
(-1)^2-4(1)(-4) = 1-4(-4) = 1+ 16 = 17 and it's not a perfect square.. dahhhh
anonymous
  • anonymous
What is the point of intersection?
CoconutJJ
  • CoconutJJ
|dw:1435379165728:dw|
UsukiDoll
  • UsukiDoll
you graph the two equations first. and then they should have common points.
CoconutJJ
  • CoconutJJ
|dw:1435379200509:dw|
UsukiDoll
  • UsukiDoll
with 0 being one of them
anonymous
  • anonymous
they're not nice numbers. about (6.56, 2.56)
anonymous
  • anonymous
|dw:1435379227291:dw|
CoconutJJ
  • CoconutJJ
So if we use the shell method from 0 to 4, we are basically using the disc method?
UsukiDoll
  • UsukiDoll
double integrals.. double the anti derivatives :S
anonymous
  • anonymous
|dw:1435379315446:dw|
anonymous
  • anonymous
\[ \int dV = 2\pi\int rh~dy = 2\pi\int \limits_0^{y_1} y(f(y)-g(y))~dy \]
CoconutJJ
  • CoconutJJ
THANK YOU! I think I can do it from here now :)
anonymous
  • anonymous
The volume of a cylinder is: \[ V = \pi r^2h \]When you differentiate, it is: \[ \frac{dV}{dr} = 2\pi r h \implies \int dV = 2\pi\int rh~dr \]That's how I think about it.
CoconutJJ
  • CoconutJJ
Thankks

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