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Alchen [17]
3 years ago
5

How can you find asymptote for the function y=x /4 + 16/x

Mathematics
1 answer:
zhannawk [14.2K]3 years ago
5 0
Note that the first term of this function has no asymptote, as there is no division by zero.  On the other hand, 16/x is undefinited at x=0.  Thus, the vertical asymptote is x=0.

Were you asked to consider horiz. asy. as well?  This function has none, since y=x/4 continues to grow; its slope is 1/4.


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Set up a double integral for calculating the flux of the vector field ????⃗ (????⃗ )=????⃗ , where ????⃗ =⟨x,y,z⟩, through the p
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Answer:

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Step-by-step explanation:

F (r) = r = (x, y, z) the surface equation z = 3(x^2 + y^2) z_x = 6x, z_y = 6y the normal vector n = (- z_x, - z_y, 1) = (- 6x, - 6y, 1)

Thus, flux ∫∫s F · dS is given as;

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=∫∫ <x, y, 3x² + 3y²> · <-6x, -6y, 1>dA , since z = 3x² + 3y²

Thus, flux is;

= ∫∫ -3(x² + y²) dA.

Since the region of integration is bounded by x² + y² = 25, let's convert to polar coordinates as follows:

∫(θ = 0 to 2π) ∫(r = 0 to 5) -3r² (r·dr·dθ)

= 2π ∫(r = 0 to 5) -3r³ dr

= -(6/4)πr^4 {for r = 0 to 5}

= -(6/4)5⁴π - (6/4)0⁴π

= -937.5π

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