Integration. About -ve area and stationary value.

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Greetings to all. :smile:

Okay, I have 2 simple questions about integration. They are summarized in topic title.

1) This is the figure:
http://usera.ImageCave.com/biosyn/qq.PNG

It shows a curve y=x^2 and the line y=3+2x.
Point A is (3,9) and point B is (-1,1).

We have to calculate the area of the region enclosed between the curve and the line segment AB.

I have used the formula A=[tex]\int[/tex] (y2-y1) dx. Correct me if the formula is wrong. I've used the limits 3 and -1. Now, I have solved this and obtained the answer, but the problem is that the Area I'm getting is coming in negative. The answer to this question is 32/3 which is what I'm getting but only that it is -32/3.

Is this something to do with the figure or have I done something wrong?

2) This is the second part of a question. First part is done. We were asked to show the Volume, V cm^3, was given by:
V=12[tex]\pi[/tex]r2-2[tex]\pi[/tex]r3.

Okay, now the second part of the question says that Given that r varies, find the stationary value of V.

What I did here is, because the stat.value says that dV/dx = 0, I used:
0(dV/dx)=12[tex]\pi[/tex]r2-2[tex]\pi[/tex]r3.
0=24[tex]\pi[/tex]r-6[tex]\pi[/tex]r2.

After this, I don't know what else to do. The answer is coming as 64[tex]\pi[/tex]. But how is this coming?

Any help? ::confused:
 

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  • #2
LCKurtz
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Greetings to all. :smile:

Okay, I have 2 simple questions about integration. They are summarized in topic title.

1) This is the figure:
http://usera.ImageCave.com/biosyn/qq.PNG

It shows a curve y=x^2 and the line y=3+2x.
Point A is (3,9) and point B is (-1,1).

We have to calculate the area of the region enclosed between the curve and the line segment AB.

I have used the formula A=[tex]\int[/tex] (y2-y1) dx. Correct me if the formula is wrong. I've used the limits 3 and -1. Now, I have solved this and obtained the answer, but the problem is that the Area I'm getting is coming in negative. The answer to this question is 32/3 which is what I'm getting but only that it is -32/3.

Is this something to do with the figure or have I done something wrong?

Did you use -1 for the lower limit and 3 for the upper limit or did you have them backwards?

2) This is the second part of a question. First part is done. We were asked to show the Volume, V cm^3, was given by:
V=12[tex]\pi[/tex]r2-2[tex]\pi[/tex]r3.

What volume? Was this formula given or is it something you calculated?

Okay, now the second part of the question says that Given that r varies, find the stationary value of V.

What I did here is, because the stat.value says that dV/dx = 0, I used:
0(dV/dx)=12[tex]\pi[/tex]r2-2[tex]\pi[/tex]r3.
0=24[tex]\pi[/tex]r-6[tex]\pi[/tex]r2.

After this, I don't know what else to do. The answer is coming as 64[tex]\pi[/tex]. But how is this coming?

Any help? ::confused:

So you would factor it:

[tex]6\pi r(4 - r^2) = 0[/tex]

and r = 2 is the only reasonable value giving a non-zero volume. What volume to you get if r = 2? I'm guessing you may have a sign error in your volume calculation, which you didn't show.
 

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