find the volume of the solid bounded above the surface z=f(x,y) and below the pl
ID: 2848322 • Letter: F
Question
find the volume of the solid bounded above the surface z=f(x,y) and below the plane region R. Please show work
Find the volume of the solid bounded above by the surface z = f(x,y) and below by the plane region R: f(x, y) = 2x + y; R is the triangle bounded by y = 2x, y = 0 and x = 2. f(x, y) = ex+2y; R is the triangle with vertices (0,0), (1,0), and (0,1). Find the average value of the function f(x, y) = xy over the plane region bounded by y = x, y = 2 - x. and y = 0. Find the maximum and minimum values of the function f(x, y) = ex-y subject to the constraint x2 + y2 = 1.Explanation / Answer
The volume is 4 times the volume between the first quadrant
in the x-y plane and the surface. The part in the x-y plane is the
quarter circle x^2+y^2=1.
The volume in the first quadrant is
Sdx from x=0 to 1. S from x to ?(1-x^2) of (1-x^2-y^2)dy
where A denotes the integral
=Sdx.[(1-x^2)y - (y^3)/3] from x to ?(1-x^2)
=Sdx((2/3)(1-x^2))^(3/2) from x=0 to x=1
Sub x=sinu, dx=cosu du and limits from u=0 to ?/2
giving
S(2/3)cos^4(u) du
=(2/3)(3/4)(1/2)(?/2) using the quick method
=?/8
The answer is 4 times this so
?/2
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