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.

- CoconutJJ

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- schrodinger

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- 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

|dw:1435377786774:dw|
This is the bounded area

- CoconutJJ

Wait so how does the integral change when you rotate it around the x axis instead of y?

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## More answers

- misssunshinexxoxo

https://www.wyzant.com/resources/lessons/math/calculus/integration/finding_volume

- misssunshinexxoxo

With "washer method"

- 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

Follow rules of "shell" and "washer" method to guide throughout finding volume

- 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

and then the limits will be the y coordinates of the intersection points

- CoconutJJ

so then the thickness of the cylinder is dy?

- CoconutJJ

|dw:1435378522268:dw|

- UsukiDoll

@peachpi it's been a while for me too.

- misssunshinexxoxo

Peachpi seems absolutely right; this is a tough area

- anonymous

this is a solid. so there's no hole, right?

- anonymous

the area lies on the axis it's being rotated on

- 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

well they intersect when x=6.562

- anonymous

yeah, but the integral is in y.
a is 0.
b you solve yÂ² = y + 4 to get

- CoconutJJ

How is the integral with respect to y ?

- 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

oh I see where the a and b for the integral is |dw:1435378968903:dw|

- UsukiDoll

those intersection points are the values for a and b

- CoconutJJ

|dw:1435378996664:dw|

- 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 " yÂ² = y + 4 to get"

- anonymous

I think it is simpler to do it horizontally than vertically, which would require two integrals.

- UsukiDoll

hmmm y^2-y-4 = 0 and then factor ?
OH NO! I DO NOT WANT TO DO 2 INTEGRALS

- UsukiDoll

(-1)^2-4(1)(-4) = 1-4(-4) = 1+ 16 = 17 and it's not a perfect square.. dahhhh

- anonymous

What is the point of intersection?

- CoconutJJ

|dw:1435379165728:dw|

- UsukiDoll

you graph the two equations first. and then they should have common points.

- CoconutJJ

|dw:1435379200509:dw|

- UsukiDoll

with 0 being one of them

- anonymous

they're not nice numbers. about (6.56, 2.56)

- anonymous

|dw:1435379227291:dw|

- CoconutJJ

So if we use the shell method from 0 to 4, we are basically using the disc method?

- UsukiDoll

double integrals.. double the anti derivatives :S

- anonymous

|dw:1435379315446:dw|

- anonymous

\[
\int dV = 2\pi\int rh~dy = 2\pi\int \limits_0^{y_1} y(f(y)-g(y))~dy
\]

- CoconutJJ

THANK YOU! I think I can do it from here now :)

- 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

Thankks

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