Question #15bff

1 Answer
Morgan
Jun 14, 2017

See below.

Explanation:

When finding the volumes of solids of revolution, we can think about the area under a given curve as being divided up into rectangles, like we learned when we were first taught about integrals.

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If we imagine revolving each of those rectangles about the x-axis, we obtain circular disks which approximate the solid.

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Each rectangle revolves into a cylinder. Formally stated, for #a<=x_k<=b#, the rectangle with height #|f(x_k)|# and base of length #Deltax=(b-a)/n#, revolves into a disk which is a right circular cylinder having radius of base #|f(x_k)|# , height of #Deltax# and volume #pi{f(x_k)}^2Deltax# #units^2#.

We can therefore find the volume of the entire solid by adding up each individual volume of the cylinders. So for a solid of revolution we have:

#sum_(k=1)^npi{f(x_k)}^2Deltax# ~ #V=int_a^bpi{f(x)}^2dx#

Where #f(x)=4/(x^2+4)#.

The first thing we can do to get a better visualization of what this solid should look like is to graph the given curves.

Here is a graph of the area bounded by the above curve, #x=0#, and #x=2#:

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Since we are revolving this area around the x-axis, we will integrate from #x=0# to #x=2#. This is the length along which we are summing up the volumes of the cylinders. This gives:

#int_0^2pi(4/(x^2+4))^2dx#

Evaluating this integral will prove to be the more difficult part of the problem.

We can start by simplifying

#16piint_0^2(1/(x^2+4)^2)dx#

Now we will use trigonometric substitution.

#x=2tan(theta)#

#dx=2sec^2(theta)d theta#

Substituting back into our integral:

#16piint1/((2tan(theta))^2+4)^2*2sec^2(theta)d theta#

Note that I have left out the limits of integration as we have changed variables, meaning that we would need to write our limits in terms of #theta# instead of #x#. Since we will be switching back to #x# by the time we are ready to evaluate the integral, I won't bother doing the conversion.

We can simplify the integral further:

#16piint((2sec^2(theta))/(16tan^4(theta)+32tan^2(theta)+16))d theta#

#=>16piint((sec^2(theta))/(8tan^4(theta)+16tan^2(theta)+8))d theta#

#=>16piint((sec^2(theta))/(8(tan^4(theta)+2tan^2(theta)+1)))d theta#

#=>2piint((sec^2(theta))/(tan^4(theta)+2tan^2(theta)+1))d theta#

#=>2piint((sec^2(theta))/(tan^2(theta)+1)^2)d theta#

Use the trigonometric identity: #tan^2(theta)+1=sec^2(theta)#

#=>2piint((sec^2(theta))/(sec^2(theta))^2)d theta#

#=>2piint1/(sec^2(theta))d theta#

#=>2piintcos^2(theta)d theta#

Use Pythagorean Identity: #cos^2(theta)=1/2(1+cos(2theta))#

#=>2piint1/2(1+cos(2theta))d theta#

#=>piint(1+cos(2theta))d theta#

Split the integral:

#piint1d theta+piintcos(2theta)d theta#

The left is a simple integral, yielding #pitheta#. For the right, we can solve using a basic #u# substitution.

#u=2theta#

#du=2d theta#

Which gives that #d theta=1/2du#. Substituting back into the integral:

#=>pitheta+pi/2intcos(u)du#

#=>pitheta+pi/2sin(u)#

Substituting back in that #u=2theta#:

#=>pitheta+pi/2sin(2theta)#

Now we go back to where we used the original trig substitution, #x=2tantheta#. Solving for #theta# gives #theta=arctan(x/2)#. We can substitute this back into the integral:

#=>pi(arctan(x/2))+pi/2sin(2*arctan(x/2))#

And then we can evaluate using our original limits of integration:

#pi(arctan(x/2))+pi/2sin(2*arctan(x/2))|_0^2#

Which gives

#piarctan(1)+pi/2sin(2arctan(1))#

#=>pi(pi/4+1/2)#

#~~4.0382#

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