Chapter 8 Solutions

These are explanations that I had emailed to various students while we were working on Chapter 8.  Use this information as you review old problems, or while you're studying for the final exam.

Remember, my Google Drive folder contains copies of worksheets and ANSWERS (on the last page of each file). CLICK HERE TO ACCESS.

Chp 8 Review #2 wkst:

See my work on the second and third pages here:

Chp 8 Review (pg. 427) Rs and Ts:

R5b) Ok, here's the "big idea" process for rotation of area problems:

STEP 1: Draw a picture of the situation on your paper.  Seriously, it will really help on these! :)

When you draw the cross-sections on this diagram, they should be horizontal "washers", where the center of both circles is on the y-axis.  The edge of the outer circle touches y = x, and the edge of the inner circle touches y = x^0.25.

STEP 2: Decide whether your problem should be in terms of x or in terms of y.

This shape is being rotated around the y-axis, so our equation should be in terms of y (we need to solve for x).  Our boundaries will also be y-axis boundaries.

Change y = x^0.25 to:

y^4 = x

The other equation doesn't really need to be changed, because it's y = x.

STEP 3: Find the area of one cross-section (usually a circle or a "washer" where it's a big circle with a small circle subtracted from the middle).

See how the radius of each circle is horizontal?  In other, the radius of each circle is an x-length out to the curve?  So, we need to use whatever "x" is equal to for the "radius" in our circle area equations.

Area of the "big circle":

pi * radius^2

pi * y^2

Area of the "little circle":

pi * radius^2

pi * (y^4)^2

pi * y^8

Area of one "washer" cross-section is "big circle" minus "little circle":

pi * y^2 - pi * y^8

STEP 4: Find the boundaries (where do the cross-sections start, and where do they end?)

Draw (or imagine) a lot more cross-sections drawn on your picture.  They would stack up around the y-axis, where they'd start at the bottom y-value (y = 0) and end up at the top y-value (y = 1).  I found these numbers by setting the two equations equal to each other and solving for y:

y = y^4

0 = y^4 - y

0 = y(y^3 - 1)

y = 0 and y = 1

So, 0 and 1 will be the integral boundaries, because we are trying to add up all the cross sections in between those two values.  (Also, these boundaries are in terms of the y-axis).

STEP 5: Integrate the equation for the area of one cross-section from the beginning boundary to the ending boundary.

(integral sign from 0 to 1) (pi*y^2 - pi*y^8) dy

Use your calculator or integrate by hand.  You should get the same answer either way. Hope this helps!

T6) Because we’re rotating around the y-axis, we need all equations in terms of y (in other words, solve for x). We started with y = x^3, which becomes x = y^(1/3).

We need to write an equation for the area of one cross section. Here, because of the rotation, our cross sections will be the shape of circles, so we’ll be finding the area of a circle: pi*r^2.

The radius of this circular cross-section is the distance from the y-axis to the curve. This is why we needed to change our equation to x = y^(1/3). This distance from the y-axis to the curve is a horizontal length, or in other words is the distance of the x-value of the function.

This makes the radius of our circular cross section: y^(1/3)

This makes the area of one circular cross section: pi(y^(1/3))^2

We’re going to integrate this area equation. Our equation is in terms of y, so we’ll use “dy”, and our bounds will be y-values:

integral (from 0 to 8) pi(y^(1/3))^2 dy

You can move the pi to the outside of the integral sign, because it’s a constant, and you can simplify the exponents:

pi * integral (from 0 to 8) y^(2/3) dy

Then integrate! Make the exponent one larger, then divide by the new exponent:

pi * 3/5 * y^(5/3)

Then plug in 8, then plug in 0, then subtract. The answer should be 19.2pi or 60.319.

"Chp 8 + Misc AP Review Problems" #1 worksheet: 

2) A point of inflection only occurs when the concavity CHANGES. So, set f’’(x) = 0, find each possible point of inflection, then check the concavity on either side of it by plugging values into f’’(x) that are chosen from either side of the x-values you just found. In this problem, f’’(x) = 0 when x = 0, 3, 6. So, check the values on either side of 0, 3, and 6. For example, you could find f’’(-1), f’’(1), f’’(4), and f’’(7)

You can pick any numbers to plug into f’’(x), as long as they’re on either side of your values. It’s only an actual point of inflection if the concavity is positive on one side then negative on the other side of the value (or vice versa). 

5) This is asking you to find the derivative of an integral, which would just leave you with the equation inside the integral (called the “integrand”). Here, F’(x) = sqrt(x^3 + 1). You only need to plug in 2 to this equation to find the final answer. It’s an example of the second form of the Fundamental Theorem of Calculus.

If you did this the long way (finding the integral, then finding the derivative, then plugging x = 2 into the derivative), it would look like this:

(anti-derivative with x in it) - (anti-derivative with 0 in it)

Then you’d find the derivative of this to get F’(x). The first part would return to your original equation, but with x as the variable instead of t. The second part would go away, because it’d be the derivative of a constant. So, all you’re left with is the original equation, which you’d then plug x = 2 into. 

6) Find the derivative f’(x). Don’t forget the chain rule.

f’(x) = 2*sec^2 (2x)

Then plug in pi/6.

f’(pi/6) = 2*sec^2 (2*pi/6)

2*sec^2 (pi/3)

2 / cos^2 (pi/3)

2 / (1/2)^2 ………… this step shows that cos (pi/3) = 1/2

2 / (1/4)

2 * 4

8

7) To find the equation of a tangent line, you need the slope and you need a point. They already gave you the slope, which is 1. We can use the information they gave us to solve for the point.

If they told us that f’(x) = 1, we should find f’(x), then set it equal to 1. We can then solve for x.

f’(x) = 4x^3 + 4x

1 = 4x^3 + 4x

To solve for x, you could graph both sides of this equation, then find their intersection points. Or, you could make the equation equal zero, then use your calculator’s Solver function. You should find that x = 0.2367. Plug this x-value back into f(x) to find the y-value of the point we need.

f(0.2367) = .2367^4 + 2(.2367)^2 = 0.1152

Now, you have slope = 1 and you have the point (0.2367, 0.1152) for your tangent line. You could write your tangent line equation in point-slope, then change it to slope-intercept.

y - 0.1152 = 1(x - 0.2367)

y - 0.1152 = x - 0.2367

y = x - 0.122

8) The wording of this problem is the same as saying that F(x) is equal to the integral of the given equation. So, integrate the given equation to find F(x):

You could use u-substitution, where u = ln x and du = 1/x * dx

This makes it: u^3 * du

The integral of this would be: 1/4 u^4 + c

Which is really: 1/4(ln x)^4 + c

They gave you F(1) = 0, which means that when you plug x = 1 into your integrated equation, it should be equal to 0. This will let us solve for the +c.

0 = 1/4(ln 1)^4 + c

0 = 1/4(0)^4 + c

0 = c

So, our particular solution for F(x) = 1/4(ln x)^4

Now find F(9) by plugging in x = 9:

F(9) = 1/4(ln 9)^4 = 5.827

Sec 8.5 Volume of Solids and Slabs GENERAL ADVICE:

We have to think about WHERE the shape is rotating.

If it rotates around the x-axis, we want to integrate the area of each cross section in terms of x.

If it rotates around the y-axis, we want to integrate the area of each cross section in terms of y.

Sec 8.5 (pg. 390):

#1a) It says this shape is rotating about the y-axis.  So, we need our equation in terms of y (in other words, solve for x).

y = 9-x^2

y-9 = -x^2

9-y = x^2

sqrt(9-y) = x

This is the radius of each circular cross section.  Look at the picture…the radius is the length of the x-value (horizontal). So, the area of each circular cross section will be:

pi * radius^2 = pi * sqrt(9-y)^2) = pi * (9-y)

Now, we want to add up all those circular cross sections from y = 0 to y = 9 (notice, our boundaries are referring to the Y-AXIS).  So, we want to integrate the equation for the area of one cross section from 0 to 9:

(integral sign) pi * (9-y)dy

pi (integral sign) (9-y)dy

Now use the FTC, and integrate with respect to y:

pi * (9y - 1/2y^2)

Now plug in 9, then plug in 0, then subtract:

pi * (9*9 - 1/2*9^2) - pi * (9*0 - 1/2*0^2)

pi* (81 - 81/2) - 0

pi * 40.5

127.235

#1b)  Integrating numerically means USE YOUR CALCULATOR and check your work above.

#1c)  This shape is kind of like a cone (but not really) and kind of like a cylinder (but not really).  It should be bigger than a cone with the same dimensions, but smaller than a cylinder with the same dimensions.  So, find the area of a cone, and the area of a cylinder with the same dimensions, and confirm that your rotated shape falls right in between those two answers.  This is one way to confirm geometrically.

#1c continued) You'll need to solve for the intersection of y = 9 - x^2 and the x-axis to find the radius of the "cone" and "cylinder".  Then, the intersection of y = 9 - x^2 and the y-axis will provide the height of the cone and cylinder.

The height of all shapes on the y-axis is 9.

The curve intersects the x-axis at x = 3 and x = -3, so the radius of the shapes is 3.

Volume of cylinder: pi * 3^2 * 9 = pi * 81

Volume of cone: 1/3 * pi * 3^2 * 9 = pi * 27

#2b) Same idea for this problem as 1c.  The intersection of the x-axis and the line y = 10 - 2x will give you the radius of the "cone" and "cylinder".

y = 10 - 2x = 0

10 = 2x

5 = x

So, the radius of both shapes is 5.

#5) It says the shape is in Quadrant 1, and is bounded by y = 1 and y = ln x.  This makes it so the area doesn't "go on forever".  The first step in ALL of these problems is to DRAW THE GRAPH ON YOUR PAPER.  Trust me, it'll help! :)

Now that you have the area for the graph drawn, try drawing the rotated cross-section.  Because it's rotating around the y-axis, it should be a horizontal circle where the radius of the circle goes from the y-axis to the curve y = ln x.  See how the radius is horizontal?  The length of the radius is the x-value of the y = ln x curve.  So, if we want to find the area of this circular cross section, we need an equation for the radius first:

y = ln x 

Solve this for x, to get it in terms of y:

e^y = x

Now you have an equation for the x-length from the y-axis out to the curve, because your equation is now "x=".  This new equation is your radius!  So, the area of one circular cross section is:

pi * radius^2

pi * (e^y)^2

pi * e^(2y)

Great!  Now we have the area of one cross section.  To find the whole volume, we need to "add up" all those circular cross sections from y = 0 up to y = 1.  Let's integrate the equation for the area of one cross-section:

(integral sign from 0 to 1) pi * e^(2y)

Answer = 10.036

BIG IDEA: When you're finding a shape rotated around the X-AXIS, your equation needs to be in terms of x, with horizontal x-axis boundaries.

When you're finding a shape rotated around the Y-AXIS, your equation needs to be in terms of y, with vertical y-axis boundaries.

#6) Because this shape is created by rotating around the Y-AXIS, you need everything in terms of y.  Start by solving your equation for x, to put it in terms of y:

y = x^(3/4)

y^(4/3) = x

Hopefully, you have already drawn a picture for this situation.  It WILL help, even more than just graphing it on your calculator, because you can sketch an idea of what the cross-section looks like.  Since your shape is rotating around the y-axis, you should have drawn a circular horizontal cross-section, where the center is on the y-axis and the outer edge of the circle touches your line x = y^(4/3).

We need to find the area of one circular cross-section, which means we need the radius of that circle.  See how the radius is horizontal, starts at the y-axis, and ends at the line?  Because it's horizontal, the radius length is the x-value of the line.  That's why we solved the equation for x, so that we could figure out what x was equal to!

So, the radius = y^(4/3)

So, the area of one circular cross-section is:

pi * (y^(4/3))^2

= pi * y^(8/3)

To find the volume of the whole shape, we need to use y-axis boundaries, because that shows where it starts and stops, and also everything else in our problem has been in terms of y.  The problem stated that we are bounded by y = 1 and y = 8.  We will integrate the equation for the area of one circular cross-section:

(integral sign from 1 to 8) pi * y^(8/3)

Answer: 1753.865

#7) This one is tough!  First off, DRAW A PICTURE.  It'll help, I promise!  Once you have the picture on your paper, try drawing the circular cross-section as it rotates around the y-axis.  It should be horizontal, with a horizontal radius.  Because it's a horizontal shape (because it rotates around the y-axis), we need all equations and boundaries to be in terms of y.

Solve for x in both of your equations, to put them in terms of y:

y = 8x   becomes…    x = 1/8y

y = x^4   becomes…   x = y^(1/4)

Your horizontal cross-section should look like a "washer".  The outer circle has a center on the y-axis, and touches the x = y^(1/4) curve.  The inner circle has a center on the y-axis, and touches the x = 1/8y line.  We need the area of both circles, because we'll want to subtract the smaller circle from the larger circle.

Area of large circle:

pi * radius^2

pi * (y^(1/4))^2

pi * y^(1/2)

Area of small circle:

pi * radius^2

pi * (1/8y)^2

Area of "washer" is large circle - small circle:

pi * y^(1/2) - pi * (1/8y)^2

Great, we have the area of one cross section!  To find the entire volume, we need to integrate the equation for one cross-section, from the first y-axis boundary to the last y-axis boundary.  We're using y-axis boundaries because we're rotating around the y-axis.  To find the boundaries, find the two y-values where the curves cross:

Find y-boundaries:

1/8y = y^(1/4)

y = 8y^(1/4)

y^(3/4) = 8

y = 8^(4/3)

y = 16  (Also y = 0, because there was one point in this calculation where we divided by y, which took away one of the solutions.  You can also graph on your calculator to find the points where they cross.)

Area of rotated shape:

(integral sign from 0 to 16) pi * y^(1/2) - pi * (1/8y)^2

I'll leave the FTC calculations up to you.  When you're done, check your answer with fnInt on your calculator.

Answer: 67.021 in^3

#13) It will be REALLY helpful to draw a picture of this shape before trying to figure out the integral.  Even though it's shown in the book, it will help to draw it yourself, label the pieces you need, etc.

In this problem, the shape is rotating around the vertical line x = 3.  Because it's a rotation around a vertical line, it's kind of like rotating around the y-axis, so your equations and boundaries used need to be in terms of y.

Look at the picture you've drawn.  One cross section should be a "washer", where the outer circle has a center at x = 3 and goes out to the y-axis on the left side and x = 6 on the right side.  The inner circle has a center at x = 3, and goes out to y = 4-x^2 on the left side.

Before we start anything else, we need to convert our y = 4-x^2 into terms of y instead of terms of x:

y = 4 - x^2

y - 4 = -x^2

-y + 4 = x^2

4 - y = x^2

sqrt(4 - y) = x

To find the area of one cross section, we need "Area of large circle - Area of small circle".

Area of large circle: pi * radius^2, where the radius is from x = 3 to the y-axis (a distance of 3).

= pi * 3^2

= pi * 9

Area of small circle: pi * radius^2, where the radius is from x = 3 to the curve x = sqrt(4 - y)

= pi * (3 - sqrt(4 - y))^2

Area of one cross section:

= pi * 9 - pi * (3 - sqrt(4 - y))^2

Now, for integrating to find the volume, we need to integrate the equation for the area of one cross section from the "beginning" of the shape to the "end" of the shape.  We will use the y-axis boundaries, because this is a rotation around a vertical line.  The shape starts at y = 0 and ends at y = 4.

Integrating to find the volume:

(integral sign from 0 to 4) pi * 9 - pi * (3 - sqrt(4 - y))^2 dy

Use fnInt on your graphing calculator.  Answer = 75.398