Ok Any Math Wise Help

[BreakABone]

Argh! I hate missing school!

anyhow I ended up missing Friday and like that day of notes. I'm now trying to make up the homework but got stuck on a bound.

I'm supposed to find the area of a graph bounded by the y-axis.

The equation is

y= square root of (16-x²)

The graph bis bounded on the y-axis between 0 and 4

How exactly does this work?

I can do X-Axis..I was just wondering if you switch it from dy/dx to dx/dy or something

[Jason1]

*runs off to get Xantar*

[Xantar]

I'm not sure what your problem with dy/dx or dx/dy is. But in any case, the equation is a graph for a semi-circle with radius 4 sitting entirely above the x-axis. So to find the area, just find the area of a circle with radius 4 and divide by 2.

There might be a way to integrate that kind of equation, but I never bothered.

[BreakABone]

Originally posted by Xantar
I'm not sure what your problem with dy/dx or dx/dy is. But in any case, the equation is a graph for a semi-circle with radius 4 sitting entirely above the x-axis. So to find the area, just find the area of a circle with radius 4 and divide by 2.

There might be a way to integrate that kind of equation, but I never bothered.

Actually in order to do this problem correctly you must integrate...

I can do the integration of it no problem

y'=16-x²
y=16x-x³/3

That isn't the real issue here.. The graph is bounded by y..So I'm wonderinf it the problem needs to be in terms of y..and if so,.. do you have to change the whole problem

[Xantar]

Is this another case of your teacher insisting that you use a more complicated method than necessary? :hmm:

If the graph were bounded on the y-axis from, say, 0 to 3, I'd agree with you. But the graph is bounded from 0 to 4. Plug in any value you want for x and you'll see that the y-value never goes above 4. The boundaries set by the problem are meaningless because the graph never goes beyond those boundaries anyway.

Oh, and by the way, your integration is wrong. It's the correct integration for 16 - (x squared), but you forgot that the entire equation is under a square root.

[BreakABone]

Originally posted by Xantar
Is this another case of your teacher insisting that you use a more complicated method than necessary? :hmm:

If the graph were bounded on the y-axis from, say, 0 to 3, I'd agree with you. But the graph is bounded from 0 to 4. Plug in any value you want for x and you'll see that the y-value never goes above 4. The boundaries set by the problem are meaningless because the graph never goes beyond those boundaries anyway.

Oh, and by the way, your integration is wrong. It's the correct integration for 16 - (x squared), but you forgot that the entire equation is under a square root.

Actually the integration is correct.. you don't know the full formula for the area of a region bounded by either axis

It's

Pi times the integration of {R(X)}²dx

So I skipped the step where I took out the square..

As a square root squared would give you what ever was inside the sign.. I assumed you knew that much

And actually that is the way to do iyt.. there is also a pic in the book

[GameMaster]

What math course is this?

[Xantar]

Okaaay...

If your integration is correct, I still don't see why you have to integrate in the first place. The graph is a semi-circle with the center at the origin and a radius of 4. And it doesn't go beyond the boundaries set by the question.

So what's the problem?

[BreakABone]

Originally posted by Xantar
Okaaay...

If your integration is correct, I still don't see why you have to integrate in the first place. The graph is a semi-circle with the center at the origin and a radius of 4. And it doesn't go beyond the boundaries set by the question.

So what's the problem?

The problem I stated in the second post

I need to know if I change the equation into terms of y or leave it in terms of x. You are finding the region bounded in terms of y but the original equation is in terms of X.

[Xantar]

I told you that the boundaries don't matter. But I think I finally understand what you're asking me. The answer to your question is no, you don't need to state the equation in terms of y. You can integrate the entire thing from x = -4 to x = 4. If the boundaries had been on the y-axis from 0 to 3, I would say that you should rewrite the equation. But in this case, you don't have to.

I still don't see why you're integrating, but I guess if you really think there's a reason, there must be. I've never pretended to remember everything from my calculus classes.

[BreakABone]

Originally posted by Xantar
I told you that the boundaries don't matter. But I think I finally understand what you're asking me. The answer to your question is no, you don't need to state the equation in terms of y. You can integrate the entire thing from x = -4 to x = 4. If the boundaries had been on the y-axis from 0 to 3, I would say that you should rewrite the equation. But in this case, you don't have to.

I still don't see why you're integrating, but I guess if you really think there's a reason, there must be. I've never pretended to remember everything from my calculus classes.

Like I said before..I'm just learning this stuff.. and just missed this lesson..So I have to do it by the books.. The books give me a formula.. I use the formula...

What can I say

And the graph is bounded by y from 0 to 4 and according to my textbook it does go pass those boundaries....

Damnit I hate misplacing my Graphing calc

[ed328]

I'll assume that when you say bounded by the y-axis, you mean bounded on the left. You've also stated that the function is valid for only a positive y. Therefore all we have to do is integrate the function from x=0 to x=4.

The integral of sqrt(16-x²) = 1/2*x*sqrt(16-x²) + 8*Arcsin(x/4)

At x=4, this is 8*(PI/2) = 4PI
At x=0, this is 8*0 = 0

4PI-0=4PI

The area is 4 PI.

You could also just take the area of the circle with radius 4 divided by 4 which would give you the same answer.

[Bizzyman]

i can stand on my head for 5! seconds


WOOOOOOOOOOOOOOOOOP!

[BreakABone]

Well if anyone cares to work it out..

The answer is 128pi/3

[ed328]

Originally posted by BreakABone
Well if anyone cares to work it out..

The answer is 128pi/3

That is the answer if y=PI*(16-x²).

But y=sqrt(16-x²) which does not equal PI*(16-x²).

What exactly are the conditions that you have to use your formula PI*integral of [R(x)]²dx? From the information you've given us, Xantar and I are correct.

[BreakABone]

Originally posted by ed328


That is the answer if y=PI*(16-x²).

But y=sqrt(16-x²) which does not equal PI*(16-x²).

What exactly are the conditions that you have to use your formula PI*integral of [R(x)]²dx? From the information you've given us, Xantar and I are correct.

Yeah I realized where I made the mistake.. you are supposed to find the volume not the area.. oops...