# A Calculus 3 question

1. Dec 3, 2006

### marc.morcos

Hey guys, i need some help... im 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 cant seem to get the limits of integration... when i sketch it it doesnt quite make sense.... thx in advance...

2. Dec 4, 2006

### marc.morcos

i tried again but the sketch doesnt make perfect sense, its almost like the boundaries dont limit it completly, 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 dont see the use of being given the x=y plane because its just skims the parabolic cylinder, not giving it a boundary...

3. Dec 4, 2006

### HallsofIvy

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
$$\int_{x=0}^{1/4}\int_{y= x}^{(1/2)\sqrt{x}}\int_{z= 0}^x z dzdydx$$

4. Dec 4, 2006

### marc.morcos

thx so much... i couldnt see how the x=y would effect

5. Dec 4, 2006

### marc.morcos

oh and i think that z goes from 0 to 5x not 0 to x