Question

The region between the graphs of y=x^2 and y=5x is rotated around the line y=25.
Find the volume

Answer 1

so far i have integral 0 to 25 ((5-sqrt(x)^2)-(5-(x/5))^2

Answer 2

The axis is y=25, and I don't see you subtracting anything from 25.

You can also opt to use the washer method, if that's what you're doing at school.

Answer 3

so how do i solve this can u do it step by step?

Answer 4

Do you prefer the shell method or washer method?

Answer 5

Have you done either of the two methods in your course?

Answer 6

washer

Answer 7

There you go, then the integral would be the difference of two squares.
The first one is the "outer radius", meaning from the axis to the outmost part of the area x^2 in our case. This is (25-x^2)^2.
The second square is from the axis to the edge of the "hole", i.e. empty space we don't want to integrate, in our case this is (25-5x)^2.
Multiply all by pi to get:
\[\pi \int\limits_{0}^{5}((25-x^2)^2-(25-5*x)^2)dx\]
Does that look familiar?

Answer 8

how do u get 25-5x

Answer 9

Notice that we are integrating along the x-direction, so dx would represent a vertical slice.

(25-5x) is the distance between the axis of rotation and the top of our region. This represents the "hole" of the washer, or the area that we need to subtract from a disk (to make a washer.

Answer 10

OOOHHH

Answer 11

yes ok

Answer 12

Find the volume of the solid obtained by rotating the region bounded by the given curves about the line x=−4
y=x^2, x=y^2

Answer 13

how about something like that using shell, now this i have no idea how to set up using shell but i was advised to

Answer 14

?

Answer 15

Sorry, I'm back!

Answer 16

Find the volume of the solid obtained by rotating the region bounded by the given curves about the line x=−4
y=x^2, x=y^2

Answer 17

how about something like that using shell, now this i have no idea how to set up using shell but i was advised to

Answer 18

This one you can do either shell or washer.

Answer 19

You could start with making a plot to find the points of intersection.

Answer 20

ok

Answer 21

give me a sec

Answer 22

0 to 1?

Answer 23

Yep!
I suggest you do this by the method of washers to practice what we just did. If you succeed, then we can use the shell method as a check.

Answer 24

ok lets do it

Answer 25

Can you figure out the outer and inner radius?

Answer 26

tha was wat i was just about to ask

Answer 27

y=x^2 is outter

Answer 28

Even before that, can you tell me which functions you used to find the intersection?

Answer 29

There is x^2, is the other one sqrt(x)?

Answer 30

Sqrt(x) is the part of x=y^2 above the x-axis. We like to do that because x=y^2 is not strictly a function.

Answer 31

Are you still there?

Answer 32

yes

Answer 33

ok i understand

Answer 34

So the axis is at y=-4, right?

Answer 35

yes

Answer 36

Can you give me the "outer" radius, meaning the expression that gives the distance of the top part of the area relative to the axis (y=-4, which is below the area).

Answer 37

i think its stil y=x^2

Answer 38

x^2 is the lower curve. From the axis (y=-4), sqrt(x) is the outer (upper) curve. Do you think so?

Answer 39

1 second

Answer 40

yes i saw it once i zoomed in

Answer 41

sqrt

Answer 42

of x

Answer 43

So the distance for the "outer" radius is therefore sqrt(x)+4, right?

Answer 44

yes

Answer 45

And the "inner" radius (referrng to the lower curve x^2) is then....

Answer 46

so (sqrt(x)-4)-x^2

Answer 47

Not really, for the inner, we're just interested in the x^2 curve, so it would be x^2+4

Answer 48

(sqrt(x)+4)-(x^2+4)

Answer 49

Almost there, you need to square them, and multiply everything by pi.
This way, pi(outer)^2 is the big disk, and pi(inner)^2 is the small (empty) disk.
The difference becomes a washer that you want to calculate.

Answer 50

yes ok, let me put it into my program to see wat i get, give me a minute

Answer 51

Fantastic it was right

Answer 52

i have one more but i think i know wat the upper an lower radius' are

Answer 53

Give it a try!

Answer 54

Find the volume of the solid obtained by rotating the region bounded by the curves
y=x^4, y=1

Answer 55

about ?

Answer 56

outter radius is 4-x^2?

Answer 57

inner is 4-1?

Answer 58

Is it about y=+4?

Answer 59

yes

Answer 60

x^2 or x^4?

Answer 61

Find the volume of the solid obtained by rotating the region bounded by the curves
y=x^4, y=1
about the line y=4

Answer 62

What you suggested seem OK to me, just change the x^2 to x^4, must have been a typo.

Answer 63

wait why x^4?

Answer 64

I thought your question asked for x^4, right?

Answer 65

oh yes lol

Answer 66

my bad

Answer 67

no problem! lol

Answer 68

i got it wrong though

Answer 69

i got 248/45pi

Answer 70

Can you show me what you put in?

Answer 71

\[\int\limits_{0}^{1}(4-x^4)^2-(4-1)^2)\]

Answer 72

Let met see...

Answer 73

It's just the pi missing, unless the answer does not correspond to the question.

Answer 74

Let's review the question:
Ah, what are your limits?

Answer 75

x^4 is an even function!

Answer 76

0 to 1

Answer 77

I suppose you have not yet graphed the function to find the enclosed area?

Answer 78

An even function typically take - to +, like -1 to +1.

Answer 79

ok so what did u get as the interval?

Answer 80

-1 to +1, that's where the two functions intersect.

Answer 81

yes u were right

Answer 82

So everything's good?

Answer 83

yes thanks a lot

Answer 84

Let's resume:
1. find the shape of the area to rotate, and find the limits (graphing).
2. if using the washer method, find the outer and inner radii.
3. integrate using the correct limits, and don't forget pi.
OK?

Answer 85

hahaha but i finished my homework professor

Answer 86

That's OK, you'll still have exams to come, that's when YOUR professor has the last laugh! lol

Answer 87

Anyway, have a great weekend, and enjoy!

Answer 88

thanks a lot

Answer 89

You're welcome!

Answer 90

when are u usually on?

Answer 91

Evenings.

Answer 92

ok catch u later

Answer 93

Yep!