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Arturiano [62]
3 years ago
9

Find the volume of the following solid. The solid between the cylinder ​f(x,y)equals=e Superscript negative xe−x and the region

Requals=StartSet left parenthesis x comma y right parenthesis : 0 less than or equals x less than or equals ln 6 comma negative 6 less than or equals y less than or equals 6 EndSet{(x,y) : 0≤x≤ln6, −6≤y≤6}. An x y z coordinate system has a positive x-axis with a tick mark at ln 6, a y-axis with tick marks at negative 6 and 6, and a positive z-axis. A surface labeled f left parenthesis x comma y right parenthesis equals e Superscript negative x has a rectangular projection in the x y plane that extends between x-coordinates 0 and ln 6 and between y-coordinates negative 6 and 6. All cross sections of the surface parallel to the x z plane are identical curves that fall at a decreasing rate from x-coordinate 0 to a x-coordinate ln 6, with a z-coordinate greater than 0 at x-coordinate ln 6. ln 6ln6 66 66 The volume of the solid is
Mathematics
1 answer:
PilotLPTM [1.2K]3 years ago
7 0

It is hard to comprehend your question. As far as I understand:

f(x,y) = e^(-x)

Find the volume over region R = {(x,y): 0<=x<=ln(6), -6<=y <= 6}.

That is all I understood. It would be easier to understand with a picture or some kind of visual aid.

Anyways, to find the volume between the surface and your rectangular region R, we must evaluate a double integral of f on the region R.

\iint_{R}^{ } e^{-x}dA=\int_{-6}^{6} \int_{0}^{ln(6)} e^{-x}dx dy

Now evaluate,

\int_{0}^{ln(6)}e^{-x}dx

which evaluates to,  5/6 if I did the math correct. Correct me if I am wrong.

Now integrate this w.r.t. y:

\int_{-6}^{6}\frac{5}{6}dy = 10

So,

\iint_{R}^{ } e^{-x}dA=\int_{-6}^{6} \int_{0}^{ln(6)} e^{-x}dx dy = 10

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Suppose a population of rodents satisfies the differential equation dP 2 kP dt = . Initially there are P (0 2 ) = rodents, and their number is increasing at the rate of 1 dP dt = rodent per month when there are P = 10 rodents.  

How long will it take for this population to grow to a hundred rodents? To a thousand rodents?

Step-by-step explanation:

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\frac{dp}{dt} =kp^2\\\\1=k(10)^2\\\\k=\frac{1}{100}

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100=-\frac{200}{2t-100} \\\\2t-100=-\frac{200}{100} \\\\2t=98\\\\t=49\ months

You have 1000 rodents when:

1000=-\frac{200}{2t-100} \\\\2t-100=-\frac{200}{1000} \\\\2t=99.8\\\\t=49.9\ months

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