<p>A hole of radius r is bored through the center of a sphere of radius R (greater than) r. Find the volume if the remaining portion of the sphere when r = 6 and R = 10. </p>
<p>Can anyone help me solve this?!</p>
<p>Work out the volume of the sphere between the limits, then work out the volume of the cylinder(hole). Then subtract.</p>
<p>hmmm rainbow kirby, thats what the problem looks like by intuition, however, the piece that is cut out, the "cylinder," isnt actually a cylinder. its actually a cylinder with slightly rounded tops. of course, the hole left in the sphere would look like a true cylinder, but the solid taken out would be different.</p>
<p>I think you should use integrals to solve....</p>
<p>this problem has nothing to do with calculus..</p>
<p>its just a basic solid geometry problem</p>
<p>(4/3)<em>pi</em>100 - (4/3)<em>pi</em>216</p>
<p>yeah it doesnt need calculus, but its not as straightforward as variance thinks.</p>
<p>The problem is that the top and bottom surfaces will be curved, so the volume of the bored out hole will be slightly larger than the volume of a cylinder with the same radius. That's where the calculus part comes in.</p>
<p>If the radius of R = 10, then the equation of the circle is
X^2 +y^2 = 100
the equation of the semicircle is y = root(100-x^2)
find the values of x (they will be 2, but they'll just be opposites, I'll call the negative value A and positive B) for which y = 6 which is the radius of the hole</p>
<p>use disks to find the volume of the shape created when you rotate the region from x=-10 to A [or from x=B to x=10] about the x axis.
multiply that value by two
add it to the volume of a cylinder with radius 6 and height of 2B (or absolute value of A +B)
subtract that from the volume of a sphere with radius 10.</p>
<p>it's most certainly an AB Calc problem.</p>
<p>and I'm not sure how you'd do it without calculus...meh.</p>
<p>Hmm... calculus. My calculus is a bit rusty</p>
<p>I'd say we picture a coordinate system. Now, plot a semi-circle with origin at (0,0). Are you given numbers for this? Well, I'll presume not, let's see...</p>
<p>The equation of a circle with origin at 0,0 is: R^2 = x^2 + y^2, solving for y and then taking only the positive half (it's easier than the negative half ;)) you get Y = sqrt(R^2-X^2). This can be your f(x).</p>
<p>Next find the line that is r away from the X axis (this will be the radius of the hole you will bore through the sphere)... which is simply Y = r. this will be your g(x).</p>
<p>Now, do a typical CALC AB Volumes of solids of revolution using dx. Looking down the hole you will have slices of circles with holes cut out in them. But now, these areas are easy to find. Since f(x) serves as your outside circle radius, do the (pi)[f(x)]^2 thing with that and subtract the inside hole, which will be (pi)[g(x)]^2. Then integrate the whole thing with respect to x with bounds from x = -R to +R (since it is simply the radius of the big sphere.</p>
<p>and VOILA.</p>
<p>EDIT: HHAHA sorry, I'm a fool, you are given numbers for this, the same still applies, jsut plug in numbers for the variables. </p>
<p>If anyone finds any error in this (I'm tired) feel free to correct!</p>
<p>LadyinRed- the above posters misread the problem. I'm pretty sure they think that the hole is a concentric sphere completely encompassed by the larger sphere... which I do not believe it is.</p>
<p>Wait wait... I realized the bounds are not going to be -R to R... just kidding. They will be the intersections between f(x) and g(x), which is pretty easily found.</p>
<p>no, but the same thing applies, if a cylinder is bore through the middle u'd just integrate the volume of the sphere for the the big one... and then subtract out the int[y=r] from -R to R... thus solving the problem!</p>
<p>in response to pebbles' third post:</p>
<p>i don't think you can do it that way...because now you're not taking into account those extra curved out ends...you need at least two integral expressions to represent the bored out hole. One for those curved end sections and on for the cylinder shape. :?</p>
<p>I'm not bothering with the cut-out sections, just the parts that are intact. ;)</p>
<p>we're using completely different approaches</p>
<p>cujoe - integrating y=r would give you a flat area, not a volume. And even if you were to integrat (pi)r^2, you'd be cutting out a cylinder from a sphere with flat ends, and a bigger cylinder than possible with a certain radius.</p>
<p>yes...we're using completely different approaches...</p>
<p>but only integrating pi r^2 [regardless of limits of integration] for the volume of the hole doesn't give the right answer because it's essentially just subtracting a cylinder.</p>
<p>I'm not subtracting anything but a circle ;)</p>
<p>It could very possibly still be wrong, but I want to make sure you actually understand me correctly :)</p>
<p>This seems like a rather straightforward volumes by washers problem. I think my solution is similar to the ones above but not identical.</p>
<p>The equation for the semicircle is y=sqrt(100-x^2).
And the hole would act as the line y=6.
Find the two points of intersection 6=sqrt(100-x^2), these would be the limits of integration, a and b.
Then integrate from a to b of pi<em>[sqrt(100-x^2)^2-6^2] which simplifies to the integral from a to b of pi</em>(64-x^2).</p>
<p>Aw, don't do all his work FOR him... guidance, not answers ;)</p>
<p>I just realized my solution is identical to pebbles solution, after he/she revised it by saying the bounds are intersections, not -10 to 10. Oh well...it's easier to look at it when it's all in one post.</p>
<p>I get it now.</p>
<p>HH23 - both methods are right. Use whatever's quicker and easier. That's most likely pebbles' way, unless you're like me and the shorter way always confuses you....</p>