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Volumes of Revolution (Discs & Washers Method)

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Volumes of Revolution by Cross Sections (Washers)
While we commonly use integrals to find area under curves. If we imagine the graph "coming out of the page" and revolving in 3-D, we can extend our concepts of calculating area to calculating Volumes of Revolution. 

Volumes of Solids by Cross Sections


The volume VVV of a solid between x=ax=ax=a and x=bx=bx=b having cross-sectional area A(x)A(x)A(x)is rotated about the x-axis is 
V=∫abA(x)dx\boxed{V=\int_a^bA(x)dx}V=∫ab​A(x)dx​
The cross-section of the solid generated in the plane perpendicular to the x-axis.

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Volumes of Solids by Washers
If your cross-sectional area is a disk, then the area of that disk will be πr2\pi r^2πr2. For a functionf(x)f(x)f(x)the cross-sectional area isA(x)=π[f(x)]2A(x)=\pi[f(x)]^2A(x)=π[f(x)]2. 


For a function f(x)f(x)f(x), the volume of the solid of revolution obtained by revolving the region betweenf(x)f(x)f(x)and the x-axis between x=ax=ax=a and x=bx=bx=b  about the x-axis is given by
V=∫abπ[f(x)]2dx\boxed{V=\int^b_a\pi [f(x)]^2dx}V=∫ab​π[f(x)]2dx​
This formula is useful for revolving about lines parallel to the x-axis.
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Volumes Between Curves by Washers




If f(x)≥g(x)≥0f(x)\ge g(x)\ge0f(x)≥g(x)≥0 then the volume of the solid of revolution obtained by revolving the region between f(x)f\left(x\right)f(x) and g(x)g\left(x\right)g(x) between x=ax=ax=a and x=bx=bx=b about the x-axis is given by
V=∫abπ([f(x)]2−[g(x)]2)dx\boxed{V=\int_a^b\pi([f(x)]^2-[g(x)]^2)dx}V=∫ab​π([f(x)]2−[g(x)]2)dx​

More generally, think of the formula as
V=∫abπ([outer radius]2−[inner radius]2)dx\boxed{V=\int_a^b\pi([\text{outer\ radius}]^2-[\text{inner\ radius}]^2)dx}V=∫ab​π([outer radius]2−[inner radius]2)dx​


Watch Out!
It may be necessary to compute the points of intersection in order to obtain aaa and bbb.


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Volumes about the Y-Axis using by Washers
If we'd like to use the Washer method about the y-axis, we need to write our equation as a function of yyy.

V=∫abπ[g(y)]2dy\boxed{V=\int^b_a\pi [g(y)]^2dy}V=∫ab​π[g(y)]2dy​

V=∫abπ([f(y)]2−[g(y)]2)dy\boxed{V=\int_a^b\pi([f(y)]^2-[g(y)]^2)dy}V=∫ab​π([f(y)]2−[g(y)]2)dy​
These formulas is useful for revolving about lines parallel to the y-axis.

Wize Concept
For the method of Washer the variables in the integral should be same of the axis of rotation. (revolving about the x-axis →use dx
\rightarrow \text{use } dx→use dx, revolving about the y-axis →use dy
\rightarrow \text{use } dy→use dy) 

Watch Out!
Be careful to consider the inner and outer radii; this depends on the axis of rotation.

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Example: Volumes of Revolution 
Find the volume of the solid obtained by revolving the region between x2+1x^2+1x2+1 and 3−x3-x3−x around the xxx-axis.

We begin by finding the intersection points of the curves. Setting x2+1=3−xx^2+1=3−xx2+1=3−x, we get 0=x2+x−2=(x−1)(x+2)0=x^2+x−2=(x−1)(x+2)0=x2+x−2=(x−1)(x+2), and so the intersection points are x=−2x=−2x=−2 and x=1x=1x=1.
We have that on [−2,4][−2,4][−2,4], 3−x≥x2+13−x\ge x^2+13−x≥x2+1, so the volume of the solid is given by
V=π∫−21[3−x]2−[x2+1]2dx\displaystyle V = \pi\int^1_{−2}[3 − x]^2 − [x^2 + 1]^2 dxV=π∫−21​[3−x]2−[x2+1]2dx

=π∫−21(9−6x+x2−x4−2x2−1)dx\displaystyle= \pi\int^1_{−2}(9 − 6x + x^2 − x^4 − 2x^2 − 1)dx=π∫−21​(9−6x+x2−x4−2x2−1)dx

=π∫−21(−x4−x2−6x+8)dx\displaystyle= \pi\int^1_{−2}(−x^4 − x^2 − 6x + 8) dx=π∫−21​(−x4−x2−6x+8)dx

=π(−x55−x33−3x2+8x)∣−21\displaystyle=\left.\pi\left(−\frac{x^5}{5} −\frac{x^3}{3} − 3x^2 + 8x\right)\right|^1_{−2}=π(−5x5​−3x3​−3x2+8x)​−21​

=π[(−15−13−3+8)−(325+83−12−16)]\displaystyle= \pi\left[\left(−\frac{1}{5} −\frac{1}{3} − 3 + 8\right)−\left(\frac{32}{5}+\frac{8}{3} − 12 − 16\right)\right]=π[(−51​−31​−3+8)−(532​+38​−12−16)]

=π(−335−3+5+28)\displaystyle= \pi\left(−\frac{33}{5} − 3 + 5 + 28\right)=π(−533​−3+5+28)

=π(150−335)\displaystyle= \pi\left(\frac{150 − 33}{5}\right)=π(5150−33​)

=1175π\displaystyle=\frac{117}{5}\pi=5117​π

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Example: Volumes of Revolution
Find the volume of a solid ball with radius aaa .

The ball can be generated by rotating a half-disk. Since the radius is aaa , we can use the half-disk whose area is generated by 0≤y≤a2−x20 \leq y \leq \sqrt{a^2 - x^2}0≤y≤a2−x2​ where −a≤x≤a-a \leq x \leq a−a≤x≤a , and revolve it around the x-axis. 

Note that visually, you could use a vertical half-disk instead, but that would no longer be a function of xxx . A computation in this case can still be done, but it's easier, and more natural to do it in the way suggested initially. The volume then becomes
V=π∫−aa(a2−x2)2dx=2π∫0a(a2−x2)dxV=\pi\int^a_{-a}(\sqrt{a^2-x^2})^2dx=2\pi\int^a_0(a^2-x^2)dxV=π∫−aa​(a2−x2​)2dx=2π∫0a​(a2−x2)dx

=2π(a2x−x33)∣0a=4πa33= \left . 2\pi\left(a^2x-\frac{x^3}3\right) \right |^a_0=\frac{4\pi a^3}{3}=2π(a2x−3x3​)​0a​=34πa3​

Practice: Volumes of Revolution
The volume of the solid that is produced by revolving the region bounded between y=x2y=x^2y=x2 and y=xy=\sqrt xy=x​ about the x-axis. 

Your answer should involve

\pi

I don't know

Practice: Volumes of Revolution
Write an integral representing the volume of the solid that is produced by revolving the region bounded between y=xy=xy=x and y=3x−x2y=3x-x^2y=3x−x2 about the line y=−1y=-1y=−1. Do not compute the integral.

V=π∫13[3x−x2+1]2−(x+2)2dx\displaystyle V = \pi\int^3_1[3x − x^2 + 1]^2 − (x + 2)^2 dxV=π∫13​[3x−x2+1]2−(x+2)2dx

V=π∫02[3x−x2+1]2−(x+1)2dx\displaystyle V = \pi\int^2_0[3x − x^2 + 1]^2 − (x + 1)^2 dxV=π∫02​[3x−x2+1]2−(x+1)2dx

V=π∫02[4x−x2+1]2−(x+2)2dx\displaystyle V = \pi\int^2_0[4x − x^2 + 1]^2 − (x + 2)^2 dxV=π∫02​[4x−x2+1]2−(x+2)2dx

V=π∫13[4x−x2+1]2−(x+1)2dx\displaystyle V = \pi\int^3_1[4x − x^2 + 1]^2 − (x + 1)^2 dxV=π∫13​[4x−x2+1]2−(x+1)2dx

None of the above

I don't know