[MultiVarCalc] Find the mass of the solid

In summary, the mass of the solid bowl consisting of points inside the paraboloid z=a(x^2+y^2) for 0≤z≤H, with a mass density of z, is equal to H^3/2a.
  • #1
bornofflame
56
3

Homework Statement


Let ##a > 0##. Find the mass of the "solid bowl" consisting of points inside the paraboloid ##z=a(x^2+y^2) \text { for } 0\leq z \leq H \text{. Assume a mass density of } \rho(x, y, z) = z##.

Homework Equations


##x^2 + y^2 = z^2##

The Attempt at a Solution


math263_review5.jpg
[/B]
mass = ##m =\int_W \rho(x, y, z)\, dV = \int \int \int \rho(x,y,z) dV = \int \int \int zr\, dz\, dr\, d\theta##
boundaries: ##0 \leq z \leq H;~ 0 \leq r \leq \sqrt \frac H a;~ 0 \leq \theta \leq 2\pi##
##m = \int_0^{2\pi} \int_0^{\sqrt \frac H a} \int_0^H zr\,dz\,dr\,d\theta##
##= \int_0^{2\pi} \int_0^{\sqrt \frac H a} \frac 1 2 z^2r\left. \right|_0^H \,dr \,d\theta##
##=\frac 1 2 \int_0^{2\pi} \int_0^{\sqrt \frac H a} H^2r \,dr \,d\theta##
##= \frac 1 2 \int_0^{2\pi} \frac 1 2 H^2r^2\left. \right|_0^{\sqrt \frac H a}\, d\theta##
##= \frac 1 4 \int_0^{2\pi}| H^2(\sqrt \frac H a)^2\, d\theta##
##= \frac 1 4 \int_0^{2\pi} H^2(\frac H a)\, d\theta##
##= \frac 1 4 \int_0^{2\pi} \frac {H^3} a \,d\theta##
##= \frac 1 4 \frac {H^3} a \theta \left. \right|_0^{2\pi}##
##= \frac {H^3} {4a} 2\pi##
##= \frac {H^3} {2a}##
 

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  • #2
bornofflame said:

Homework Statement


Let ##a > 0##. Find the mass of the "solid bowl" consisting of points inside the paraboloid ##z=a(\sqrt{x^2+y^2)} \text { for } 0\leq z \leq H \text{. Assume a mass density of } \rho(x, y, z) = z##.
That equation isn't a paraboloid. It is a cone. Which do you want?

Homework Equations


##x^2 + y^2 = z^2##

The Attempt at a Solution


View attachment 225683 [/B]
mass = ##m =\int_W \rho(x, y, z)\, dV = \int \int \int \rho(x,y,z) dV = \int \int \int zr\, dz\, dr\, d\theta##
boundaries: ##0 \leq z \leq H;~ 0 \leq r \leq \sqrt \frac H a;~ 0 \leq \theta \leq 2\pi##
##m = \int_0^{2\pi} \int_0^{\sqrt \frac H a} \int_h^H zr\,dz\,dr\,d\theta##

What is ##h##? And in the step below you change it to ##0##. That isn't correct in any case.

##= \int_0^{2\pi} \int_0^{\sqrt \frac H a} \frac 1 2 z^2r\left. \right|_0^H \,dr \,d\theta##
##=\frac 1 2 \int_0^{2\pi} \int_0^{\sqrt \frac H a} H^2r \,dr \,d\theta##
##= \frac 1 2 \int_0^{2\pi} \frac 1 2 H^2r^2\left. \right|_0^{\sqrt \frac H a}\, d\theta##
##= \frac 1 4 \int_0^{2\pi}| H^2(\sqrt \frac H a)^2\, d\theta##
##= \frac 1 4 \int_0^{2\pi} H^2(\frac H a)\, d\theta##
##= \frac 1 4 \int_0^{2\pi} \frac {H^3} a \,d\theta##
##= \frac 1 4 \frac {H^3} a \theta \left. \right|_0^{2\pi}##
##= \frac {H^3} {4a} 2\pi##
##= \frac {H^3} {2a}##
 

Related to [MultiVarCalc] Find the mass of the solid

1. How is mass calculated for a solid object?

Mass is calculated by finding the product of the density of the material and its volume. The formula for mass is: mass = density x volume.

2. What is the unit of measurement for mass?

The unit of measurement for mass is typically kilograms (kg) in the metric system and pounds (lbs) in the imperial system.

3. Can the mass of a solid object be negative?

No, the mass of a solid object cannot be negative. Mass is a physical property and cannot have a negative value.

4. How does the shape of a solid object affect its mass?

The shape of a solid object does not affect its mass. The mass of an object is determined by its volume and the density of the material, which remain constant regardless of the shape of the object.

5. What is the difference between mass and weight?

Mass is a measure of the amount of matter in an object, while weight is a measure of the force exerted on an object by gravity. Mass is constant, but weight can vary depending on the strength of gravity at a given location.

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