Calculus Problems, triple integral and polar coordinates stuff

In summary: But I think I can do it. 2. For number two I have the integral as r^2dzdrd\theta, with the z limits 0 to 9-r^2, the r limits 0 to 3, and the \theta limits 0 to \pi/2...I am mainly uncertain about the \theta limits, but am not sure about the others or the integral either. Is it right?Yes, that's correct. The θ limit is always 0 to 2π. 3.A volcano fills the volume between the graphs z=0 and z=\frac{1}{
  • #1
mmmboh
407
0
Hi I have a homework set due this week, 14 problems, I have done 11 of them, but these 3 are giving me trouble, help would be great :)

Homework Statement



1.A cylindrical drill with radius 4 is used to bore a hole through the center of a sphere of radius 8. Find the volume of the ring shaped solid that remains.

2.Use cylindrical coordinates to evaluate the triple integral [tex]\int\int\int_{E}[/tex] [tex]\sqrt{x^2+y^2}[/tex]dV , where E is the solid bounded by the circular paraboloid z=9−(x^2+y^2) and the xy -plane.

3.A volcano fills the volume between the graphs z=0 and z=[tex]\frac{1}{(x^2+y^2)^2}[/tex], and outside the cylinder x^2+y^2=1. Find the volume of this volcano.

The Attempt at a Solution



1. For the first one I know the volume of a sphere is 4/3[tex]\pi[/tex]r^3...so I found the volume and then I subtracted what I thought would be the volume of the cylinder, [tex]\pi[/tex]r^2h...for the height I used 16 because that's the diametre of the sphere sphere but I realized that the height of the cylinder will be a bit less but I can't figure out how much, and I would like to solve it with calculus too if I can but I can't figure out how to set up the integral.

2. For number two I have the integral as r^2dzdrd[tex]\theta[/tex], with the z limits 0 to 9-r^2, the r limits 0 to 3, and the [tex]\theta[/tex] limits 0 to [tex]\pi/2[/tex]...I am mainly uncertain about the [tex]\theta[/tex] limits, but am not sure about the others or the integral either. Is it right?

3. For the third I am not sure how to set up the integral, help please.

Thank you!
 
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  • #2
Hi mmmboh! :smile:

(have a theta: θ and a pi: π and an integral: ∫ and try using the X2 tag just above the Reply box :wink:)
mmmboh said:
1.A cylindrical drill with radius 4 is used to bore a hole through the center of a sphere of radius 8. Find the volume of the ring shaped solid that remains.

With problems like this, always slice the region into slices whose volume you already know.

In this case, slice the hole (well, you know what i mean! :rolleyes:) into concentric cylindrical shells of thickness dx. :smile:
2.Use cylindrical coordinates to evaluate the triple integral [tex]\int\int\int_{E}[/tex] [tex]\sqrt{x^2+y^2}[/tex]dV , where E is the solid bounded by the circular paraboloid z=9−(x^2+y^2) and the xy -plane.

2. For number two I have the integral as r^2dzdrd[tex]\theta[/tex], with the z limits 0 to 9-r^2, the r limits 0 to 3, and the [tex]\theta[/tex] limits 0 to [tex]\pi/2[/tex]...I am mainly uncertain about the [tex]\theta[/tex] limits, but am not sure about the others or the integral either. Is it right?

Yes, that's fine, except that θ always goes from 0 to 2π. :smile:

3.A volcano fills the volume between the graphs z=0 and z=[tex]\frac{1}{(x^2+y^2)^2}[/tex], and outside the cylinder x^2+y^2=1. Find the volume of this volcano.

Same method as 1. :wink:
 
  • #3
mmmboh said:
Hi I have a homework set due this week, 14 problems, I have done 11 of them, but these 3 are giving me trouble, help would be great :)

Homework Statement



1.A cylindrical drill with radius 4 is used to bore a hole through the center of a sphere of radius 8. Find the volume of the ring shaped solid that remains.

2.Use cylindrical coordinates to evaluate the triple integral [tex]\int\int\int_{E}[/tex] [tex]\sqrt{x^2+y^2}[/tex]dV , where E is the solid bounded by the circular paraboloid z=9−(x^2+y^2) and the xy -plane.

3.A volcano fills the volume between the graphs z=0 and z=[tex]\frac{1}{(x^2+y^2)^2}[/tex], and outside the cylinder x^2+y^2=1. Find the volume of this volcano.



The Attempt at a Solution



1. For the first one I know the volume of a sphere is 4/3[tex]\pi[/tex]r^3...so I found the volume and then I subtracted what I thought would be the volume of the cylinder, [tex]\pi[/tex]r^2h...for the height I used 16 because that's the diametre of the sphere sphere but I realized that the height of the cylinder will be a bit less but I can't figure out how much, and I would like to solve it with calculus too if I can but I can't figure out how to set up the integral.
Draw a picture, looking at the picture from the side. You have a circle with a vertical rectangle taken out of the center. Draw a horizontal line from the center of the circle to one side of the rectangle and draw a line from the center of the circle to the point where that side of the rectangle crosses the circle. You now have a right triangle where the hypotenuse of the triangle has length the radius of the sphere, 8, and one leg has length the radius of the drill, 4. Use the Pythagorean theorem to find the length of the other side which is half the length of the rectangle. The only part of this problem that requires calculus is finding the volume of the "caps" on each end of the rectangle that are taken out by the drill.

2. For number two I have the integral as r^2dzdrd[tex]\theta[/tex], with the z limits 0 to 9-r^2, the r limits 0 to 3, and the [tex]\theta[/tex] limits 0 to [tex]\pi/2[/tex]...I am mainly uncertain about the [tex]\theta[/tex] limits, but am not sure about the others or the integral either. Is it right?
Projected into the xy-plane, you get the circle [itex]z= 9- (x^2+ y^2)= 0[/itex] or [itex]x^2+ y^2= 9[/itex], with radius 3. Take [itex]\theta[/itex] from 0 to [itex]2\pi[/itex], r from 0 to 3, and z from 0 to [itex]9- (x^2+ y^2)[/itex].

3. For the third I am not sure how to set up the integral, help please.

Thank you!
Draw the graphs of [itex]y= 1/x^4[/itex], x= -1, and x= 1 (vertical lines), giving a cross section of the figure. Of course, because of the symmetry, [itex]\theta[/itex] goes from 0 to [itex]2\pi[/itex]. The vertical line x= 1 intersects [itex]y= 1/x^4[/itex] at x= 1, of course, so r goes from 1 to infinity. Finally, for each[itex]\theta[/itex], r, z goes from 0 to [itex]1/r^4[/itex]. Integrate [itex]dV= r dzdrd\theta[/itex] with those limits of integration.
 
  • #4
Hm I'm not quite sure I completely understand how to draw the circle...
So far I have x^2+y^2=16 (equation for cylinder), and x^2+y^2+z^2=64 (sphere)

That means that z^2=64-16=48, and z=+/- [tex]\sqrt{48}[/tex], so the cylinder has a height 2[tex]\sqrt{48}[/tex]...from this I can find the volume of the cylinder, but how do I find the volume of the spherical caps?

Thanks for your guys' help! :)
 
  • #5
Any help please?
 
  • #6
mmmboh said:
1.A cylindrical drill with radius 4 is used to bore a hole through the center of a sphere of radius 8. Find the volume of the ring shaped solid that remains.
tiny-tim said:
With problems like this, always slice the region into slices whose volume you already know.

In this case, slice the hole (well, you know what i mean! :rolleyes:) into concentric cylindrical shells of thickness dx. :smile:
mmmboh said:
Hm I'm not quite sure I completely understand how to draw the circle...
So far I have x^2+y^2=16 (equation for cylinder), and x^2+y^2+z^2=64 (sphere)

That means that z^2=64-16=48, and z=+/- [tex]\sqrt{48}[/tex], so the cylinder has a height 2[tex]\sqrt{48}[/tex]...from this I can find the volume of the cylinder, but how do I find the volume of the spherical caps?

You have to cut the region into slices

Either slice the caps into concentric "vertical" cylindrical shells of thickness dr, or into "horizontal" discs of thickness dz.

Please try one of them.​
 

1. What is a triple integral?

A triple integral is an extension of the concept of an integral in calculus, where instead of integrating over one variable, you integrate over three variables. It represents the volume under a three-dimensional surface or region in space.

2. How do I solve a triple integral?

To solve a triple integral, you need to set up the limits of integration for each variable and then evaluate the integral using the appropriate integration techniques. This can involve converting the integral into a different coordinate system, such as cylindrical or spherical coordinates, to make the calculations easier.

3. What is the difference between polar coordinates and rectangular coordinates?

Polar coordinates are a two-dimensional coordinate system where a point is represented by its distance from the origin (r) and the angle it makes with the positive x-axis (θ). Rectangular coordinates, also known as Cartesian coordinates, use the x and y axes to locate a point in a two-dimensional plane.

4. How do I convert between polar coordinates and rectangular coordinates?

To convert from polar coordinates to rectangular coordinates, you can use the formulas x = rcos(θ) and y = rsin(θ). To convert from rectangular coordinates to polar coordinates, you can use the formulas r = √(x² + y²) and θ = tan⁻¹(y/x).

5. Why do we use polar coordinates in calculus?

Polar coordinates are often used in calculus because they can simplify complex integrals, particularly those involving circular or symmetric regions. They can also make it easier to visualize and solve problems in three-dimensional space. In some cases, it is more efficient to use polar coordinates than rectangular coordinates when solving certain types of problems.

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