Double integral volume problem

nate9519
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Homework Statement


find the volume of the solid below the plane z = 4x and above the circle x^2 + y^2 = 16 in the xy plane

Homework Equations

The Attempt at a Solution


This totally confused me. I didn't think the plane z = 4x sat above the xy plane. If that is true then there would be no solid between the two graphs. I ended up putting zero for the answer. was I right or wrong
 
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nate9519 said:

Homework Statement


find the volume of the solid below the plane z = 4x and above the circle x^2 + y^2 = 16 in the xy plane

Homework Equations

The Attempt at a Solution


This totally confused me. I didn't think the plane z = 4x sat above the xy plane.
Part of the z = 4x plane lies above the xy plane. Did you draw a sketch of the plane and the circle?
nate9519 said:
If that is true then there would be no solid between the two graphs. I ended up putting zero for the answer. was I right or wrong
 
$$\iint_R z \space dA = \iiint_V \space dV$$

$$\iiint_V \space dV = \iint_R \int_0^z \space dzdA$$

;)

Polar co-ordinates.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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