Calculus problem, finding the volume?

Question

A ball of radius 11 has a round hole of radius 5 drilled through its center. Find the Volume of the resulting solid?

I think we have to use the circle equation; y^2+x^2=r^2. Radius would be 11. It don't know whether to use the shell method or the disk method. Thanks

Answer 1

Either method will give you the same answer ☺
It's just whichever one you visualize the easiest. Use it to solve the problem, then force yourself to set up the problem the other way and solve it again. And remember that you onlt have to work out of the 1'st quadrant because, by symmetry, everything 'on the bottom' will be identical and you just have to multiply by 2 to get the total volume.

Another method you might like is this: Draw the quarter circle with radius r, then draw a straight line from x=5 until in intersects the quarter circle. Now draw a line to the y-axis and erase the portion of the circle 'above' the line. What you'll have (pretty obviously) is a 'cross section' of the sphere with the hole through it. Calculate the angle formed between the x (or r) axis and a line drawn to the top of what's left of the sphere (you'll have to calculate the point from the equation for the circle and use a bit of trig). Now define a differential area dx∙dy (and remember that dy is also r∙dΘ) and integrate that against dΦ over 2π radians to get a 'differential volume' that is in the form of an infitesimal 'ring' that passes through the cross section you've drawn. It will go 'through' the paper and have the axis of the drilled hole as it's axis of revolution.

Just another fun way to work the problem ☺


Doug

Answer 2

The volume is π/384 (cubic units).

You have to do a solid-of-revolution about the x-axis. The volume will be independent of the radius of the radius of the sphere. So no matter how large this sphere is when you start you will always have the same volume for it.

Answer 3

to find the volum of the ball, we use the equation -->
y^2 + x^2 = 11^2
y = (121 - x^2)^1/2

to find the volume of the hole, we use the equation -->
y = 5 (straight horizontal line)

volume = 2 (phi) (int from 0 to 11) [((121 - x^2)^1/2)^2 - (5)^2] dx
volume = 2 (phi) (int from 0 to 11) (121 - x^2 - 25) dx
volume = 2 (phi) (int from 0 to 11) (96 - x^2) dx
volume = 2 (phi) (38x + 1/3x^3) from 0 to 11
volume = 2 (phi) [(38(11) + 1/3(11)^3) - (38(0) + 1/3(0)^3)]
volume = 2 (phi) (418 + 443 2/3 - 0)
volume = 2 (phi) (861 2/3)
volume = 1723 1/3 (phi)

Answer 4

Well you're doing a problem where you take a cross-section and rotate it around the axis right? What does a cross-section of that new ball look like? (Hint: it looks like a regular half-circle except only evaluated from 5 to 11.) Try that.

Answer 5

find the volume of the sphere, find the volume of the the cylinder made by the drill, (i hope you can do the rest)