Help w/ eval of calc 3 problem

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In summary, the conversation discusses the steps and methods involved in solving calculus 3 problems, the use of calculators, and the real-life applications of calculus 3. The first step in evaluating a calculus 3 problem is to identify the type of problem and set up the appropriate equation. The method used depends on the type of problem and given information, and a calculator can be used to assist in calculations. If stuck, it is recommended to break the problem down and seek help from a tutor or professor. Additionally, calculus 3 has practical applications in various fields.
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weckod
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eval. double integal sprt(1+x^2+y^2)ds

S is helicoid and r(u,v) = ucos(v)i+usin(v)j+vk, with 0 <=u<=4 and 0<=v<=4pi

help please i don't know what a helicoid is!
 
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A helicoid is a three-dimensional curved surface that resembles a spiral staircase. In this problem, the helicoid is described by the parametric equation r(u,v) = ucos(v)i+usin(v)j+vk, with u and v being the parameters that define the surface.

To evaluate the double integral, we first need to set up the limits of integration. Since the problem states that 0 <= u <= 4 and 0 <= v <= 4pi, we can see that the surface is bounded by a circle with a radius of 4 in the x-y plane and extends vertically for a height of 4pi.

Next, we need to find the differential element ds, which is given by the formula ds = ||r_u x r_v|| du dv, where r_u and r_v are the partial derivatives of r with respect to u and v, respectively.

In this case, we have r_u = cos(v)i+sin(v)j+k and r_v = -usin(v)i+ucos(v)j. Therefore, ||r_u x r_v|| = sqrt(u^2+1).

Substituting this into the integral, we get:

∫∫ sqrt(1+x^2+y^2) ds = ∫∫ sqrt(u^2+1) du dv

Since the limits of integration are constants, we can take them out of the integral and evaluate them separately.

∫∫ sqrt(u^2+1) du dv = ∫0^4 ∫0^4pi sqrt(u^2+1) du dv

To evaluate this integral, we can use a substitution u = tan(t), which will give us du = sec^2(t) dt.

∫∫ sqrt(u^2+1) du dv = ∫0^4 ∫0^4pi sqrt(tan^2(t)+1) sec^2(t) dt dv

Using the trigonometric identity tan^2(t)+1 = sec^2(t), we can simplify the integral to:

∫∫ sqrt(u^2+1) du dv = ∫0^4 ∫0^4pi sec(t) sec^2(t) dt dv

Now, we can use the trigonometric identity sec(t) sec^2(t) = sec^3(t) to simplify the integral further:

 

1. What is the first step in evaluating a calculus 3 problem?

The first step in evaluating a calculus 3 problem is to identify the type of problem it is, such as finding a derivative or integral, and then set up the appropriate equation.

2. How do I know which method to use to solve a calculus 3 problem?

The method used to solve a calculus 3 problem depends on the type of problem and the given information. Common methods include using the chain rule or integration by parts.

3. Can I use a calculator to solve a calculus 3 problem?

Yes, a calculator can be used to help with the calculation of a calculus 3 problem. However, it is important to understand the concepts and steps involved in solving the problem before relying on a calculator.

4. What should I do if I am stuck on a calculus 3 problem?

If you are stuck on a calculus 3 problem, try breaking it down into smaller parts and working through each step. You can also seek help from a tutor or professor for clarification and additional guidance.

5. Can I use calculus 3 in real-life applications?

Yes, calculus 3 has many real-life applications in fields such as physics, engineering, and economics. It can be used to model and solve problems involving rates of change, optimization, and motion, among others.

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