Math Is Fun Forum

genericname

Member

Registered: 2012-05-16

Posts: 52

Having trouble with volume of revolution

Hi, I'm really confuse about how to set up the integral to find the volume. Why did they set up 2 integrals and add them together for the disc method? Also how do you go about choosing the correct values to use? I never understood that.

The graph and answer looks like this (sorry, didn't know how to type it with a keyboard):
i.imgur.com/ob4XQ.png

The question is:

The region in the first quadrant bounded by x=y^2, y=0 and x+y=2 is revolved about the x-axis.
a) Set up an integral(or integrals) to evaluate the resulting volume using the disc method. (Set up only)
b) Do the same problem using the shell method. (Set up only)

anonimnystefy

Real Member

From: Harlan's World

Registered: 2011-05-23

Posts: 16,049

Re: Having trouble with volume of revolution

Hi genericname

As I see it, you have to split the graph into two parts, as my picture shows, and use the disc method separately for both parts.

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“Here lies the reader who will never open this book. He is forever dead.
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment
The knowledge of some things as a function of age is a delta function.

genericname

Member

Registered: 2012-05-16

Posts: 52

Re: Having trouble with volume of revolution

Hi genericname

As I see it, you have to split the graph into two parts, as my picture shows, and use the disc method separately for both parts.

Why do you have to split it? Also how do you know when to split it?

anonimnystefy

Real Member

From: Harlan's World

Registered: 2011-05-23

Posts: 16,049

Re: Having trouble with volume of revolution

You have to split it because the distance doesn't change everywhere the same.

You can split it up when you notice a sudden change in the distance of the function from the x-axis.

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“Here lies the reader who will never open this book. He is forever dead.
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment
The knowledge of some things as a function of age is a delta function.

genericname

Member

Registered: 2012-05-16

Posts: 52

Re: Having trouble with volume of revolution

I have the height written down as ((y-2) - y^2)) in my notes, but aren't you supposed to subtract the top from bottom for height if it's on the y-axis? I'm confused about that. Shouldn't it be height  = ((y^2) - (y-2))?

Bob

Administrator

Registered: 2010-06-20

Posts: 10,626

Re: Having trouble with volume of revolution

hi genericname

Welcome to the forum.

There are two separate functions bounding the region.  The change occurs where the curve crosses the line.

So you need to treat this as two problems ... one function as x goes from 0 to 1   .... a second function from x = 1 to 2.

See graph.

Bob

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Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

anonimnystefy

Real Member

From: Harlan's World

Registered: 2011-05-23

Posts: 16,049

Re: Having trouble with volume of revolution

Hi Bob

That is what he doesn't understand why we're doing.

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“Here lies the reader who will never open this book. He is forever dead.
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment
The knowledge of some things as a function of age is a delta function.