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Featured 3 months ago

# 4sqrt(3)pia^3 #

We basically are being asked to calculate the volume of a spherical bead, that is, a sphere with a hole drilled through it.

Consider a cross section through the sphere, which we have centred on the origin of a Cartesian coordinate system:

The red circle has radius

# x^2+y^2=(2a)^2 => x^2+y^2=4a^2#

**Method 1 - Calculate core and subtract from Sphere**

First let us consider the volume of the entire Sphere, which has radius

# :. V_("sphere") = 4/3 pi (2a)^3 #

# " " = 4/3 pi 8a^3 #

# " " = 32/3 pi a^3 #

For the bore we can consider a solid of revolution. (Note this is not a cylinder as it has a curved top and bottom from the sphere). We will use the method of cylindrical shells and rotate about

# V = 2pi \ int_(alpha)^(beta) \ xf(x) \ dx #

Also note that we require twice the volume because we have a portion above and below the

# V_("bore") = 2pi \ int_(0)^(a) \ x(2sqrt(4a^2-x^2)) \ dx #

# " " = 2pi \ int_(0)^(a) \ 2x \ sqrt(4a^2-x^2) \ dx #

We can evaluate using a substitution:

Let

#u=4a^2-x^2 => (du)/dx = -2x#

When# { (x=0), (x=a) :} => { (u=4a^2), (u=3a^2) :}#

And so:

# V_("bore") = -2pi \ int_(4a^2)^(3a^2) \sqrt(u) \ du #

# " " = 2pi \ int_(3a^2)^(4a^2) \sqrt(u) \ du #

# " " = 2pi \ [ u^(3/2)/(3/2) ] _(3a^2)^(4a^2) #

# " " = (4pi)/3 \ [ u^(3/2) ] _(3a^2)^(4a^2) #

# " " = (4pi)/3 \ ((4a^2)^(3/2) - (3a^2)^(3/2)) #

# " " = 4/3pi \ (8a^3-3sqrt(3)a^3) #

# " " = 32/3pi a^3-4sqrt(3)pia^3 #

And so the total volume is given by:

# V_("total") = V_("sphere") - V_("bore" #

# " " = 32/3 pi a^3 - (32/3pi a^3-4sqrt(3)pia^3)#

# " " = 4sqrt(3)pia^3#

**Method 2 - Calculate volume of bead directly**

We can use the same method of a solid of revolution using the method of cylindrical shells and rotate about

# V_("total") = 2pi \ int_(a)^(2a) \ x(2sqrt(4a^2-x^2)) \ dx #

# " " = 2pi \ int_(a)^(2a) \ 2x \ sqrt(4a^2-x^2) \ dx #

We can evaluate using a substitution:

Let

#u=4a^2-x^2 => (du)/dx = -2x#

When# { (x=a), (x=2a) :} => { (u=3a^2), (u=0) :}#

And so:

# V_("total") = -2pi \ int_(3a^2)^(0) \sqrt(u) \ du #

# " " = 2pi \ int_(0)^(3a^2) \sqrt(u) \ du #

# " " = 2pi \ [ u^(3/2)/(3/2) ] _(0)^(3a^2) #

# " " = (4pi)/3 \ [ u^(3/2) ] _(0)^(3a^2) #

# " " = (4pi)/3 \ ((3a^2)^(3/2) - 0) #

# " " = (4pi)/3 \ (3sqrt(3)a^3 )#

# " " = 4sqrt(3)pia^3 # , as above

Featured 2 months ago

Let us denote the required integral by

# I = int \ (x-2)/(x(x^2-4x+5)^2) \ dx #

**Partial Fraction Decomposition**

We can decompose the integrand using partial fractions, the partial fraction decomposition will be of the form:

# (x-2)/(x(x^2-4x+5)^2) = A/x + (Bx+C)/(x^2-4x+5) + (Dx+E)/(x^2-4x+5)^2 #

# " " = {A(x^2-4x+5)^2 + (Bx+C)x(x^2-4x+5) + (Dx+E)x}/{x(x^2-4x+5)^2} #

Leading to:

# x-2 = A(x^2-4x+5)^2 + (Bx+C)x(x^2-4x+5) + (Dx+E)x #

# :. x-2 = A(x^4-8x^3+26x^2-40x+25) + (Bx+C)(x^3-4x^2+5x) + (Dx+E)x #

We can use various methods to find our unknown constants:

# "Put " x=0 => -2=A*5^2 #

# :. A = -2/25 #

Compare coefficients:

# "Coeff"(x^4) => 0 = A +B #

# :. B = 2/25 #

# "Coeff"(x^3) => 0 = -8A+C-4B#

# :. 0=16/25+C-8/25 => C=-8/25#

# "Coeff"(x^2) => 0 = 26A+5B-4C+D#

# :. 0=-52/25+10/25+32/25+D => D=10/25=2/5#

# "Coeff"(x^1) => 1 = -40A+5C+E#

# :. 1=80/25-40/25+E => E=-3/5#

Hence we can the integral as:

# I = int \ (-2/25)/x + (2/25x-8/25)/(x^2-4x+5) + (2/5x-3/5)/(x^2-4x+5)^2 \ dx#

# \ \ = -2/35 \ int \ 1/x dx + 2/25 int (x-4)/(x^2-4x+5) \ dx+ 1/5 int (2x-3)/(x^2-4x+5)^2 \ dx#

# \ \ = -2/25 \ I_1 + 2/25 \ I_2 + 1/5 \ I_3#

Where:

# I_1 = int \ 1/x dx + 2/25 \ dx#

# I_2 = int (x-4)/(x^2-4x+5) \ dx #

# I_3 = int (2x-3)/(x^2-4x+5)^2 \ dx#

Now let us take each of these three separate integrals in turn:

**Integral 1**:

The **first integral**,

# I_1 = int \ 1/x \ dx #

# \ \ \ = ln|x| #

**Integral 2**:

The **second integral**,

# I_2 = int \ (x-4)/(x^2-4x+5) \ dx #

# \ \ \ = int \ (x-4)/((x-2)^2-2^2+5) \ dx#

# \ \ \ = int \ (x-4)/((x-2)^2+1) \ dx#

Let

# I_2 = int \ (u-2)/(u^2+1) \ du#

# \ \ \ = int \ u/(u^2+1) - 2/(u^2+1)\ du#

# \ \ \ = 1/2int \ (2u)/(u^2+1) \ du - 2 int \ 1/(u^2+1)\ du#

# \ \ \ = 1/2ln|u^2+1| - 2 arctanu#

Restoring the substitution we get:

# I_2 = 1/2ln|x^2-4x+5| - 2 arctan(x-2)#

**Integral 3**:

The **third integral**,

# I_3 = int \ (2x-3)/(x^2-4x+5)^2 \ dx #

# \ \ \ = int (2x-3)/((x-2)^2+1)^2 \ dx #

Let

# I_3 = int (2u+1)/(u^2+1)^2 \ du #

# \ \ \ = int (2u)/(u^2+1)^2 \ du+ int 1/(u^2+1)^2 \ du #

Consider the integral:

# int (2u)/(u^2+1)^2 \ du #

Let

# int (2u)/(u^2+1)^2 \ du = int \ 1/w^2 \ dw #

# " " = -1/w #

Restoring the substitutions we get:

# int (2u)/(u^2+1)^2 \ du = -1/(u^2+1) #

# " " = -1/((x-2)^2+1) #

# " " = -1/(x^2-4x+5) #

And now we consider:

# int 1/(u^2+1)^2 \ du #

Let

# int 1/(u^2+1)^2 \ du = int \ 1/(tan^2theta+1)^2 \ sec^2 theta \ d theta #

# " " = int \ 1/(sec^2theta)^2 \ sec^2 theta \ d theta #

# " " = int \ 1/(sec^2theta) \ d theta #

# " " = int \ cos^2theta \ d theta #

# " " = int \ 1/2(1+cos2theta) \ d theta #

# " " = 1/2 \ int \ 1+cos2theta \ d theta #

# " " = 1/2 (theta + 1/2 sin2theta) #

# " " = 1/2 theta + 1/2 sintheta costheta#

Now,

# :. sinthetacostheta = cos^2theta * u #

# " " =1/sec^2theta * u #

# " " =1/(1+tan^2theta) * u #

# " " =u/(u^2+1) #

Restoring the substitutions we get:

# int 1/(u^2+1)^2 \ du = 1/2 arctanu+ 1/2 u/(u^2+1)#

# " " = 1/2 arctan(x-2) + 1/2 (x-2)/((x-2)^2+1)#

# " " = 1/2 arctan(x-2) + 1/2 (x-2)/(x^2-4x+5)#

Hence, we can write the third integral as:

# I_3 = int (2u)/(u^2+1)^2 \ du+ int 1/(u^2+1)^2 \ du #

# \ \ \ = -1/(x^2-4x+5)+ 1/2 arctan(x-2) + 1/2 (x-2)/(x^2-4x+5)#

**Combine Results**

Now we return back to our original partial fraction result from earlier, and replace the three integrals with the above results:

# I = -2/25 \ I_1 + 2/25 \ I_2 + 1/5 \ I_3#

# \ \ = -2/25 {ln|x|} + 2/25 {1/2ln|x^2-4x+5| - 2 arctan(x-2)} + 1/5 {-1/(x^2-4x+5)+ 1/2 arctan(x-2) + 1/2 (x-2)/(x^2-4x+5)} + c #

# \ \ = -2/25 ln|x| + 1/25 ln|x^2-4x+5| - 4/25 arctan(x-2) -1/5 1/(x^2-4x+5)+ 1/10 arctan(x-2) + 1/10 (x-2)/(x^2-4x+5) + c #

# \ \ = -2/25 ln|x| + 1/25 ln|x^2-4x+5| - 3/50 arctan(x-2) -1/5 1/(x^2-4x+5)+ 1/10 (x-2)/(x^2-4x+5) + c #

# \ \ = -2/25 ln|x| + 1/25 ln|x^2-4x+5| - 3/50 arctan(x-2) + (x-2-2)/(10(x^2-4x+5)) + c #

# \ \ = -2/25 ln|x| + 1/25 ln|x^2-4x+5| - 3/50 arctan(x-2) + (x-4)/(10(x^2-4x+5)) + c #

Featured 1 month ago

First set the equations equal to each other, and find the intersection points, then find the area between the curves.

By setting the equations equal to each other, I found that the functions intersect at

Focus on the unbounded (indefinite) integral to solve:

Use integration by parts:

Now go back to the definite integral:

Featured 3 weeks ago

Under some conditions, which are given below, letter series converges.

It dependes on properties of

**Theorem:** Let

For our purpose, your hypothesis to be true, we need the followings to be true:

- For all
#a_n>0# , #a_n# sequence is bounded (Let's say#|a_n|<\alpha# ),#\sum a_n# be convergent (Let's say its equal to#a# ).

If these two items are true, then your statement would be true. Moreover we can find an upper limit for

Let's start with

which means

If we continue in this respect we can find by induction (!),

Featured 2 weeks ago

We will solve this **Problem,** using the following **Leibnitz Theorem**

**(LT)** for **Derivative** of **Product** of two functions

Denoting the

We take,

**Enjoy Maths.!**

Featured 1 week ago

I'm guessing the question refers to the

#a = -sqrt(1/7)# , local minimum

#b = sqrt((1/7)# , local maximum

graph{(6x)/(1 + 7x^2) [-3.464, 3.464, -1.732, 1.732]}

graph{-(6(7x^2 - 1))/(1 + 7x^2)^2 [-3.464, 3.464, -1.732, 1.732]}

graph{(84x (7x^2 - 3))/(1 + 7 x^2)^3 [-2.464, 2.464, -24, 24]}

**DISCLAIMER:** *LONG ANSWER!*

Well, first of all, I think the question has a funky typo. It should say, "changes **from decreasing to increasing** at the point

Whenever a function changes from increasing to decreasing or vice versa, it exhibits either a **local** *maximum* or *minimum*.

For example:

#y = x^2# has one local minimum at#x = 0# , since the graph has only an upwards concavity.

graph{x^2 [-2.464, 2.464, -2, 2]}

#y = sinx# from#0# to#2pi# changes from increasing to decreasing at#x = pi/2# , and changes from decreasing to increasing at#x = (3pi)/2# .

graph{sinx [-0.05, 6.25, -1.2, 1.01]}

For your graph, the **first derivative** with respect to

Here, we have (using the product rule):

#(df)/(dx) = 6x cdot -1/(1 + 7x^2)^2 cdot 14x + 1/(1 + 7x^2) cdot 6#

#= (-6(14x^2))/(1 + 7x^2)^2 - (-6(1 + 7x^2))/(1 + 7x^2)^2#

#= (-6(14x^2 - (1 + 7x^2)))/(1 + 7x^2)^2#

#= -(6(7x^2 - 1))/(1 + 7x^2)^2#

So, you took the derivative correctly. Now, to find the points **find the zeroes** of this derivative (i.e. find when

#0 = -(cancel(6)^(ne 0)(7x^2 - 1))/cancel((1 + 7x^2)^2)^(ne 0)#

The denominator can never equal

#=> 7x^2 = 1#

So, your **critical x values** (for your local maxima and/or minima) are at

#=> color(green)(x) = pmsqrt(1/7) ~~ color(green)(pm 0.3780)#

Now that you know the

#color(green)(f(pm sqrt(1/7))) = (6(pmsqrt(1/7)))/(1 + 7(pmsqrt(1/7))^2)#

#= (pm6/(sqrt7))/(1 + 7 cdot 1/7) = pm3/(sqrt7)#

#~~ color(green)(pm 1.134)#

Right now, all you know is where the points

The **second derivative** tells you the *concavity*; up, down, or an inflection point. By the question wording, one is concave up and the other is concave down.

This derivative would be somewhat ugly

#(d^2f)/(dx^2) = -d/(dx)[(42x^2 - 6)/(1 + 7x^2)^2]# ,

but you should get:

#= (84x (7x^2 - 3))/(1 + 7 x^2)^3#

At the same critical

- If it is positive, the function is concave up and is at a minimum.
- If it is negative, the function is concave down and is at a maximum.
- Otherwise, the function is at an inflection point.

At

#ul((d^2f)/(dx^2)) = (84(-sqrt(1/7)) (7(-sqrt(1/7))^2 - 3))/(1 + 7 (-sqrt(1/7))^2)^3#

#= ((-84/sqrt7) (-2))/(2)^3 " "ul(> 0)#

At

#ul((d^2f)/(dx^2) < 0)#

So, the **left**-hand critical point is a **minimum** and the **right**-hand critical point is a **maximum**.

#color(blue)((-sqrt(1/7), -3/sqrt7) ~~ (-0.3780, -1.134))# (minimum)

#color(blue)((sqrt(1/7), 3/sqrt7) ~~ (0.3780, 1.134))# (maximum)

And we can check by graphing. Here is the function, followed by its first derivative, and then its second derivative.

graph{(6x)/(1 + 7x^2) [-3.464, 3.464, -1.732, 1.732]}

graph{-(6(7x^2 - 1))/(1 + 7x^2)^2 [-3.464, 3.464, -1.732, 1.732]}

graph{(84x (7x^2 - 3))/(1 + 7 x^2)^3 [-2.464, 2.464, -24, 24]

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