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

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50anshsrivastava 50anshsrivastava · Answer 1 · 2019-04-25T22:02:53+02:00

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.

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

Now evaluate,

[tex]\int_{0}^{ln(6)}e^{-x}dx[/tex]

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

Now integrate this w.r.t. y:

[tex]\int_{-6}^{6}\frac{5}{6}dy = 10[/tex]

So,

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