Solve Calculus 3 Question: Triple Integral over Parabolic Cylinder and Planes

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The discussion focuses on computing a triple integral of the function z over a region defined by a parabolic cylinder and several planes. The boundaries include the parabolic cylinder x = 4y^2, the planes z = 5x, y = x, and z = 0. The intersection points of the curves are identified as (0,0) and (1/4, 1/4), which help define the enclosed area. The correct limits of integration are established as x ranging from 0 to 1/4, y from y = x to y = (1/2)√x, and z from 0 to 5x. The integral is thus expressed as ∫ from 0 to 1/4 ∫ from y = x to (1/2)√x ∫ from 0 to 5x z dz dy dx.
marc.morcos
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Hey guys, i need some help... I am trying to compute the triple integral z dx dy dz, where R is the region bounded by the parabolic cylinder x= 4y^2 and the planes z = 5 x, y = x, z = 0 ... i can't seem to get the limits of integration... when i sketch it it doesn't quite make sense... thanks in advance...
 
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i tried again but the sketch doesn't make perfect sense, its almost like the boundaries don't limit it completely, so the volume appears to be infinitly large
i have the z=5x plane, x=y plane, z=0 plane and the parabolic cylinder x=4y^2... i don't see the use of being given the x=y plane because its just skims the parabolic cylinder, not giving it a boundary...
 
x= 4y2 is a parabola and y= x is a line in the xy-plane. They intersect at (0,0) and(1/4, 1/4) and inclose a small area between them. The region they include is the cylinder having those boundaries. z= 0 is the bottom boundary and z= x is the top. Overall, x ranges between 0 and 1/4. For each x, y ranges between y= x and y= (1/2)\sqrt{x}. For every (x,y), z ranges between 0 and x. Your integral is
\int_{x=0}^{1/4}\int_{y= x}^{(1/2)\sqrt{x}}\int_{z= 0}^x z dzdydx
 
thx so much... i couldn't see how the x=y would effect
 
oh and i think that z goes from 0 to 5x not 0 to x
 

Thread 'Proving that convexity implies second order derivative being positive'

There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...
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