Calculating Mass for a Cylinder in the First Octant

In summary, the part of the cylinder x2+z2=25 which is in the first octant and contained within the cylinder x2+y2=16 has a mass of 22/15.
  • #1
cp255
54
0
If the mass per unit area of a surface is given by ρ=xy, find the mass if S is the part of the cylinder x2+z2=25 which is in the first octant and contained within the cylinder x2+y2=16.So here was my attempt.

I parametrized the curve.
x2+z2=25
r(u, v) = <5cos(u), v, 5sin(u)>

I then plugged into the bounds
x2+y2=16.
25cos2(u) + v2 = 16
v = sqrt(16 - 25cos2(u))

Next I took the cross product of ru X rv. Its magnitude is a constant 5.

Now I solved the integral with bounds 0 < u < Pi/2 and 0 < v < sqrt(16 - 25cos2(u))
∫∫25 v cos(u) du dv

This returns the result of -25/3 but since we are looking for a mass I submitted the answer of positive 25/3 and this is wrong. I checked the integration on my calculator and it gets the same result.
 
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  • #2
cp255 said:
If the mass per unit area of a surface is given by ρ=xy, find the mass if S is the part of the cylinder x2+z2=25 which is in the first octant and contained within the cylinder x2+y2=16.


So here was my attempt.

I parametrized the curve.
x2+z2=25
r(u, v) = <5cos(u), v, 5sin(u)>

I then plugged into the bounds
x2+y2=16.
25cos2(u) + v2 = 16
v = sqrt(16 - 25cos2(u))

Next I took the cross product of ru X rv. Its magnitude is a constant 5.

Now I solved the integral with bounds 0 < u < Pi/2 and 0 < v < sqrt(16 - 25cos2(u))
∫∫25 v cos(u) du dv

This returns the result of -25/3 but since we are looking for a mass I submitted the answer of positive 25/3 and this is wrong. I checked the integration on my calculator and it gets the same result.

##u## doesn't go from ##0## to ##\pi/2##. Look at a picture showing what angle ##u## represents.
 
  • #3
I redid my bounds and now I have arccos(sqrt(16-v^2)/5) < u < pi/2 and 0 < v < 4 which results in the answer of 22/15 which is still wrong. I fell like I found the wrong value for u.
 
  • #4
I get 22/3 both that way and working the integral in reverse order so I think you have a mistake somewhere. But haven't you left out multiplying by the multiplier ##|r_u\times r_v|##? That would give 110/3 by my calculations. Is that the given answer?
 

Related to Calculating Mass for a Cylinder in the First Octant

1. What is a surface integral problem?

A surface integral problem is a type of mathematical problem that involves calculating the flux, or flow, of a vector field across a two-dimensional surface. It is an important concept in multivariable calculus and is often used in physics and engineering to solve problems related to fluid flow, electric and magnetic fields, and other physical phenomena.

2. How do you set up a surface integral problem?

To set up a surface integral problem, you first need to define the surface over which you want to calculate the flux. This can be done using a parametric equation or by describing the surface as a function of two variables. Next, you need to determine the vector field that is acting on the surface. Finally, you need to choose an appropriate surface element and set up the integral using the flux formula.

3. What is the significance of solving surface integral problems?

Solving surface integral problems allows us to calculate important physical quantities, such as the amount of fluid passing through a surface or the electric flux through a closed surface. These calculations are crucial in understanding and predicting the behavior of physical systems and can also be used to solve real-world engineering problems.

4. What are some common applications of surface integrals?

Surface integrals are used in a variety of fields, including fluid dynamics, electromagnetism, and computer graphics. In fluid dynamics, they are used to calculate the flow rate of a fluid through a surface, while in electromagnetism, they are used to calculate the electric and magnetic flux through a surface. In computer graphics, surface integrals are used to calculate lighting and shading effects on 3D objects.

5. Are there any techniques for solving surface integral problems?

Yes, there are several techniques for solving surface integral problems, including the parametric method, the divergence theorem, and the Stokes' theorem. These techniques rely on different mathematical concepts, such as parametric equations, partial derivatives, and curl, to simplify the integral and make it easier to solve. It is important to choose the most appropriate method for a specific problem to obtain an accurate solution.

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