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.

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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
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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
|dw:1435377786774:dw| This is the bounded area
Wait so how does the integral change when you rotate it around the x axis instead of y?

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https://www.wyzant.com/resources/lessons/math/calculus/integration/finding_volume
With "washer method"
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.
Follow rules of "shell" and "washer" method to guide throughout finding volume
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|
and then the limits will be the y coordinates of the intersection points
so then the thickness of the cylinder is dy?
|dw:1435378522268:dw|
@peachpi it's been a while for me too.
Peachpi seems absolutely right; this is a tough area
this is a solid. so there's no hole, right?
the area lies on the axis it's being rotated on
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 ?
well they intersect when x=6.562
yeah, but the integral is in y. a is 0. b you solve y² = y + 4 to get
How is the integral with respect to y ?
maybe it has something to do with the x axis being rotated so we can't use that? I have no idea x.x
oh I see where the a and b for the integral is |dw:1435378968903:dw|
those intersection points are the values for a and b
|dw:1435378996664:dw|
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"
I think it is simpler to do it horizontally than vertically, which would require two integrals.
hmmm y^2-y-4 = 0 and then factor ? OH NO! I DO NOT WANT TO DO 2 INTEGRALS
(-1)^2-4(1)(-4) = 1-4(-4) = 1+ 16 = 17 and it's not a perfect square.. dahhhh
What is the point of intersection?
|dw:1435379165728:dw|
you graph the two equations first. and then they should have common points.
|dw:1435379200509:dw|
with 0 being one of them
they're not nice numbers. about (6.56, 2.56)
|dw:1435379227291:dw|
So if we use the shell method from 0 to 4, we are basically using the disc method?
double integrals.. double the anti derivatives :S
|dw:1435379315446:dw|
\[ \int dV = 2\pi\int rh~dy = 2\pi\int \limits_0^{y_1} y(f(y)-g(y))~dy \]
THANK YOU! I think I can do it from here now :)
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.
Thankks

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