Volume Question (Can someone check my work?)

In summary, the volume of the solid within the given sphere and cone can be found by setting up a double integral in polar coordinates, with the limits of integration being 0 to 2pi for theta and 0 to sqrt(2) for r in the first integral, and 0 to 2pi for theta and sqrt(2) to 2 for r in the second integral. Solving this integral yields a volume of 8pi*sqrt(2)/3. While there is another way to solve this using a single integral, the method used here is the simplest and most straightforward.
  • #1
amcavoy
665
0
I posted this in the Calculus section also, so I apologize for double-posting.

"Find the volume of the solid that lies within the sphere [tex]x^2+y^2+z^2=4[/tex], above the x-y plane, and below the cone [tex]z=\sqrt{x^2+y^2}[/tex]."

I set this up in polar coordinates as follows:

[tex]V=\int_{0}^{2\pi}\int_{0}^{\sqrt{2}}r^2drd\theta+\int_{0}^{2\pi}\int_{\sqrt{2}}^{2}r\sqrt{4-r^2}drd\theta[/tex]

and then solved it coming up with [tex]V=\frac{8\pi\sqrt{2}}{3}[/tex].

Is this the correct way to set this up? Also, is there a way I could do this with one double integral rather than adding two of them together?

Thanks for your help.
 
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  • #2
apmcavoy said:
I posted this in the Calculus section also, so I apologize for double-posting.

"Find the volume of the solid that lies within the sphere [tex]x^2+y^2+z^2=4[/tex], above the x-y plane, and below the cone [tex]z=\sqrt{x^2+y^2}[/tex]."

I set this up in polar coordinates as follows:

[tex]V=\int_{0}^{2\pi}\int_{0}^{\sqrt{2}}r^2drd\theta+\int_{0}^{2\pi}\int_{\sqrt{2}}^{2}r\sqrt{4-r^2}drd\theta[/tex]

and then solved it coming up with [tex]V=\frac{8\pi\sqrt{2}}{3}[/tex].

Is this the correct way to set this up? Also, is there a way I could do this with one double integral rather than adding two of them together?

Thanks for your help.

It is correct, and I think that was the simplest way to solve.

ehild
 
  • #3


I cannot confirm if your calculations are correct without seeing your full work. However, your setup using polar coordinates looks appropriate for finding the volume of the solid in question. As for using one double integral instead of two, it is possible but may require more complex calculations. It would depend on your approach and preference. You could try using cylindrical coordinates instead of polar coordinates and see if it simplifies the integral. Additionally, you could also try using the triple integral in spherical coordinates. Overall, your approach using polar coordinates seems valid.
 

1. What is volume?

Volume is a measure of the amount of space occupied by an object or substance. It is typically measured in cubic units such as cubic meters or cubic centimeters.

2. How do you calculate volume?

The formula for calculating volume depends on the shape of the object. For a regular-shaped object, such as a cube or rectangular prism, the volume can be calculated by multiplying the length, width, and height. For irregular-shaped objects, such as a sphere or cylinder, there are specific formulas to calculate their volume.

3. Can you check my volume calculations?

As a scientist, I am unable to check individual volume calculations. However, there are various online tools and calculators available that can assist in verifying volume calculations.

4. How is volume different from mass and density?

Mass is a measure of the amount of matter in an object, while volume is a measure of the space occupied by that matter. Density, on the other hand, is the mass per unit volume of a substance. In other words, it is a measure of how compact or spread out the matter is within a given volume.

5. Why is volume an important concept in science?

Volume is an important concept in science because it helps us understand the physical properties of objects and substances. It is also crucial in various scientific fields, such as chemistry, physics, and engineering, as it is used to calculate and measure quantities and to make predictions and observations about the behavior of matter.

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