W01 Notes

Volume using cylindrical shells

Review

• Volume using cross-sectional area
• Disk/washer method - 01
• Disk/washer method - 02
• Disk/washer method - 03

Shells

• Shell method - 01
• Shell method - 02
• Shell method - 03

01 Theory

Take a graph in the first quadrant of the -plane. Rotate this about the -axis. The resulting 3D body is symmetric around the axis. We can find the volume of this body by using an integral to add up the volumes of infinitesimal shells, where each shell is a thin cylinder.

The volume of each cylindrical shell is :

In the limit as and the number of shells becomes infinite, their total volume is given by an integral.

Volume by shells - general formula

In any concrete volume calculation, we simply interpret each factor, ‘’ and ‘’ and ‘’, and determine and in terms of the variable of integration that is set for .

Shells vs. washers

Can you see why shells are sometimes easier to use than washers?

02 Illustration

Example - Revolution of a triangle

01 - Revolution of a triangle

A rotation-symmetric 3D body has cross section given by the region between , , , and is rotated around the -axis. Find the volume of this 3D body.

Solution Define the cross section region.

Bounded above-right by .

Bounded below-right by .

These intersect at .

Bounded at left by .

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Define range of integration variable.

Rotated around -axis, therefore use for integration variable (shells!).

Integral over :

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Interpret .

Radius of shell-cylinder equals distance along :

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Interpret .

Height of shell-cylinder equals distance from lower to upper bounding lines:

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Interpret .

is limit of which equals here so .

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Plug data in volume formula.

Insert data and compute integral:

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Exercise - Revolution of a sinusoid

02 - Revolution of a sinusoid

Consider the region given by revolving the first hump of about the -axis. Set up an integral that gives the volume of this region using the method of shells.

Solution

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Integration by substitution

[Note: this section is non-examinable. It is included for comparison to IBP.]

• Integration by Substitution 1:
• Integration by Substitution 2:
• Integration by Substitution 3:
• Integration by Substitution 4:
• Integration by Substitution 5:
• Integration by Substitution: Definite integrals, various examples

03 Theory

The method of -substitution is applicable when the integrand is a product, with one factor a composite whose inner function’s derivative is the other factor.

Substitution

Suppose the integral has this format, for some functions and :

Then the rule says we may convert the integral into terms of considered as a variable, like this:

The technique of -substitution comes from the chain rule for derivatives:

Here we let . Thus for some .

Now, if we integrate both sides of this equation, we find:

And of course .

Extra - Full explanation of -substitution

The substitution method comes from the chain rule for derivatives. The rule simply comes from integrating on both sides of the chain rule.

Setup: functions and .

Let and be any functions satisfying , so is an antiderivative of .

Let be another function and take for its independent variable, so we can write .

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The chain rule for derivatives.

Using primes notation:

Using differentials in variables:

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Integrate both sides of chain rule.

Integrate with respect to :

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Introduce ‘variable’ from the -format of the integral.

Treating as a variable, the definition of gives:

Set the ‘variable’ to the ‘function’ output:

Combining these:

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Substitute for in the integrated chain rule.

Reverse the equality and plug in:

This is “-substitution” in final form.

Integration by parts

Videos:

• Integration by Parts 1: and
• Integration by Parts 2: and
• Integration by Parts 3: Definite integrals
• Example: , two methods:
  • Double IBP
  • Fast Solution
• Integration by Parts 6:

04 Theory

The method of integration by parts (abbreviated IBP) is applicable when the integrand is a product for which one factor is easily integrated while the other becomes simpler when differentiated.

Integration by parts

Suppose the integral has this format, for some functions and :

Then the rule says we may convert the integral like this:

This technique comes from the product rule for derivatives:

Now, if we integrate both sides of this equation, we find:

and the IBP rule follows by algebra.

Extra - Full explanation of integration by parts

Setup: functions and are established.

Recognize functions and in the integrand:

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Product rule for derivatives.

Using primes notation:

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Integrate both sides of product rule.

Integrate with respect to an input variable labeled ‘’:

Rearrange with algebra:

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This is “integration by parts” in final form.

Addendum: definite integration by parts

Definite version of FTC.

Apply FTC to :

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Integrate the derivative product rule using specified bounds.

Perform definite integral on both sides, plug in definite FTC, then rearrange:

Choosing factors well

IBP is symmetrical. How do we know which factor to choose for and which for ?

Here is a trick: the acronym “LIATE” spells out the order of choices – to the left for and to the right for :

05 Illustration

Example - A and T factors

03 - A and T factors

Compute the integral:

Solution Choose .

Set because simplifies when differentiated. (By the trick: is Algebraic, i.e. more “”, and is Trig, more “”.)

Remaining factor must be :

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Compute and .

Derive :

Antiderive :

Obtain chart:

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Plug into IBP formula.

Plug in all data:

Compute integral on RHS:

Note: the point of IBP is that this integral is easier than the first one!

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Final answer is:

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Exercise - Hidden A

04 - Hidden A

Compute the integral:

Solution

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