Calculus Problems, triple integral and polar coordinates stuff

1. Jan 17, 2010

mmmboh

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 :)

1. The problem statement, all variables and given/known data

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 $$\int\int\int_{E}$$ $$\sqrt{x^2+y^2}$$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=$$\frac{1}{(x^2+y^2)^2}$$, and outside the cylinder x^2+y^2=1. Find the volume of this volcano.

3. The attempt at a solution

1. For the first one I know the volume of a sphere is 4/3$$\pi$$r^3...so I found the volume and then I subtracted what I thought would be the volume of the cylinder, $$\pi$$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$$\theta$$, 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?

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

Thank you!

Last edited: Jan 17, 2010
2. Jan 17, 2010

tiny-tim

Hi mmmboh!

(have a theta: θ and a pi: π and an integral: ∫ and try using the X2 tag just above the Reply box )
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! ) into concentric cylindrical shells of thickness dx.
Yes, that's fine, except that θ always goes from 0 to 2π.

Same method as 1.

3. Jan 17, 2010

HallsofIvy

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.

Projected into the xy-plane, you get the circle $z= 9- (x^2+ y^2)= 0$ or $x^2+ y^2= 9$, with radius 3. Take $\theta$ from 0 to $2\pi$, r from 0 to 3, and z from 0 to $9- (x^2+ y^2)$.

Draw the graphs of $y= 1/x^4$, x= -1, and x= 1 (vertical lines), giving a cross section of the figure. Of course, because of the symmetry, $\theta$ goes from 0 to $2\pi$. The vertical line x= 1 intersects $y= 1/x^4$ at x= 1, of course, so r goes from 1 to infinity. Finally, for each$\theta$, r, z goes from 0 to $1/r^4$. Integrate $dV= r dzdrd\theta$ with those limits of integration.

4. Jan 17, 2010

mmmboh

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=+/- $$\sqrt{48}$$, so the cylinder has a height 2$$\sqrt{48}$$...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. Jan 18, 2010