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

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    Calculus Calculus 3
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SUMMARY

The discussion focuses on computing the triple integral of z over the region R, which is bounded by the parabolic cylinder defined by x = 4y² and the planes z = 5x, y = x, and z = 0. The correct limits of integration are established as follows: x ranges from 0 to 1/4, y ranges from y = x to y = (1/2)√x, and z ranges from 0 to 5x. The integral is expressed as ∫(from x=0 to 1/4) ∫(from y=x to (1/2)√x) ∫(from z=0 to 5x) z dz dy dx. The discussion clarifies the boundaries and intersections of the curves involved.

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  • Familiarity with parabolic cylinders and their equations
  • Knowledge of setting limits of integration for multiple integrals
  • Ability to sketch and interpret regions in three-dimensional space
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  • Study the properties of parabolic cylinders and their applications in calculus
  • Learn how to visualize and sketch regions for triple integrals
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Students and educators in calculus, particularly those focusing on multivariable calculus, as well as anyone needing to compute triple integrals over complex regions defined by curves and planes.

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
 

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