Did I set up my integral correctly?

In summary, the conversation discusses using spherical coordinates to find the volume inside a sphere of radius 4 and outside a sphere of radius 2, as well as inside and above the cone z=3√(x^2+y^2). The solution involves replacing z^2 in the sphere of radius 4 with the cone equation and finding the radius of intersection. The integral is then taken from 0 to the calculated angle and solved using the formula ∫ 0 to 2 pi ∫0 to θ ∫2 to 4 (ρ^2sinφ) dρdφdθ. It is suggested to use LaTeX or typed solutions in the future.
  • #1
qq545282501
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1

Homework Statement


use spherical coordinates to find volume inside a sphere of radius 4 and outside sphere of radius 2, and inside/above the cone z=3√(x^2+y^2).

Homework Equations


[/B]
x=rcosθ =ρsinφcosθ , y=rsinθ =ρsinφsinθ
z=ρcosφ
r= ρsinφ

The Attempt at a Solution


replacing z^2 in the sphere of radius 4 with the cone z^2=9(x^2+y^2) I get the radius of the intersection of √( 8/5), using this radius and known ρ ( 4), φ=inverse sin of (r/ρ) i found φ to be 0.322 or around 18.4 degrees.

V= ∫ 0 to 2 pi ∫0 to 0.322 ∫2 to 4 (ρ^2sinφ) dρdφdθ or shown in the whiteboard :

Snapshot.jpg
 
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  • #2
I agree with your calculation.
 
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  • #3
qq545282501 said:

Homework Statement


use spherical coordinates to find volume inside a sphere of radius 4 and outside sphere of radius 2, and inside/above the cone z=3√(x^2+y^2).

Homework Equations


[/B]
x=rcosθ =ρsinφcosθ , y=rsinθ =ρsinφsinθ
z=ρcosφ
r= ρsinφ

The Attempt at a Solution


replacing z^2 in the sphere of radius 4 with the cone z^2=9(x^2+y^2) I get the radius of the intersection of √( 8/5), using this radius and known ρ ( 4), φ=inverse sin of (r/ρ) i found φ to be 0.322 or around 18.4 degrees.

V= ∫ 0 to 2 pi ∫0 to 0.322 ∫2 to 4 (ρ^2sinφ) dρdφdθ or shown in the whiteboard :

View attachment 91741

You should not take the integral from 0 to 0.322, but rather, take it from 0 to ## p = \arcsin \left( \frac{\sqrt{8/5}}{4} \right)##, because that will give you an exact answer in terms of a nifty formula, rather than a decimal approximation.

I future, could you PLEASE avoid posting all your really messy handwritten work (as you have done in several recent posts)? Just type it out, which you could probably do almost as quickly as you can make images of the handwritten work and upload the files, etc. This Forum greatly prefers typed solutions, although it has so far not forbidden pictures of handwriting. It would be useful to you now, and in the future, to learn how to use LaTeX; but even the type of ASCII plain text you wrote above is acceptable also.
 
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Ray Vickson said:
I future, could you PLEASE avoid posting all your really messy handwritten work (as you have done in several recent posts)?

I'm guessing the OP is creating those images with a mouse on the "whiteboard". One of the relatively recent "enhancements" of the PF site that don't make it better.
 
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  • #5
Ray Vickson said:
You should not take the integral from 0 to 0.322, but rather, take it from 0 to ## p = \arcsin \left( \frac{\sqrt{8/5}}{4} \right)##, because that will give you an exact answer in terms of a nifty formula, rather than a decimal approximation.

I future, could you PLEASE avoid posting all your really messy handwritten work (as you have done in several recent posts)? Just type it out, which you could probably do almost as quickly as you can make images of the handwritten work and upload the files, etc. This Forum greatly prefers typed solutions, although it has so far not forbidden pictures of handwriting. It would be useful to you now, and in the future, to learn how to use LaTeX; but even the type of ASCII plain text you wrote above is acceptable also.

Ok, thank you, will do. but was my handwriting that bad :frown:
 

FAQ: Did I set up my integral correctly?

What is an integral and why is it important?

An integral is a mathematical concept used to calculate the area under a curve or the volume of a 3-dimensional shape. It is important because it allows us to find solutions to a wide range of problems in fields such as physics, engineering, and economics.

How do I know if I set up my integral correctly?

One way to check if your integral is set up correctly is to use the fundamental theorem of calculus. This theorem states that the integral of a function can be calculated by finding its antiderivative and evaluating it at the upper and lower limits of integration. If the result matches the area or volume you are trying to find, then your integral is set up correctly.

What are some common mistakes when setting up an integral?

Some common mistakes include forgetting to include the correct limits of integration, using the wrong integrand, and not accounting for the correct units of measurement. It is also important to pay attention to the order of integration, as switching the order can change the result of the integral.

How can I improve my integral setup skills?

Practice is key to improving your integral setup skills. Try solving a variety of problems from different areas of math and science, and check your work using the fundamental theorem of calculus. It can also be helpful to review the properties and rules of integration.

What resources can I use to help me with setting up integrals?

There are many online resources available, such as textbooks, video tutorials, and practice problems with solutions. Your teacher or professor can also provide guidance and feedback on your integral setup. It can also be helpful to work with a study group or tutor to improve your skills.

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