A cylindrical drill with radius 5 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.
Alright, my thing is that i did not understand how to set the integral
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$\begingroup$If you Google on "hole through a sphere" or "napkin ring formula" you will find that the volume that remains when a sphere is pierced by a cylinder is dependent only on the height of the "ring".
In this case, the height of the remaining ring is $H$:$$H=2 \times (8^2-5^2)$$ Treat $\frac{H}{2}$ as the radius of a sphere, and the volume of the sphere is the answer.
Some time ago there was a loooong discussion in a math group about this problem: " I drill a 5" long hole centrally through a sphere. What is the volume of the remaining ring?"
$\endgroup$ $\begingroup$A sphere of radius $8$ is obtained by revolving a circle of radius $8$ about the $x$-axis. If you revolve the region between that circle and the line $y=5$ about the $x$-axis, you obtain the region you're interested in. That means you want to use the washer method, with $f(x)=\sqrt{64-x^2}$ and $g(x)=5$.
The limits of integration are the $x$-values of the intersection points of the graphs of $f$ and $g$.
$\endgroup$ 12 $\begingroup$First we must remember how we derive area of a sphere. We start out with $x^2 + y^2 = r^2$, or the equation of a circle. Dividing both sides by $r^2$ gives us $\frac{x^2}{r^2} + \frac{y^2}{r^2} = 1$. Subtracting both sides by $\frac{x^2}{r^2}$ gives us $\frac{y^2}{r^2} = 1 - \frac{x^2}{r^2}$. By taking the square root of both sides we have $\frac{y}{r} = \sqrt{1 - \frac{x^2}{r^2}}$ giving us a final equation of $y = r\sqrt{1 - \frac{x^2}{r^2}}$. The reason this is important is because we now have an equation for y that is dependent upon x. If you go to desmos and graph this equation, you will get a half circle with no negative y values. This should be clear when looking at the equation.
I will assume we know the area of a circle formula. If you want a proof, go to this website: . Otherwise, I will assume you know it is $\pi r^2 = A$.
By using the disc method, we determine 1. that y is the radius of each disk and 2. that, because $y = r\sqrt{1 - \frac{x^2}{r^2}}$, the area of each disk is equal to $\pi(r\sqrt{1 - \frac{x^2}{r^2}})^2$. Therefore each disk has a volume of $\pi(\sqrt{r^2}\sqrt{1 - \frac{x^2}{r^2}})$ = $\pi(\sqrt{r^2 - x^2})^2$ = $\pi r^2 - \pi x^2$. Integrating, we have
$\int_{r}^{-r}(\pi r^2 - \pi x^2 ) dx$
$(\pi r^2 x - \frac{\pi x^3}{3} ) $ between -r and r
substituting in $r$ and $-r$ for b and a, we have
$ (\pi r^2 r - \frac{\pi (r)^3}{3} )-(\pi r^2 (-r) - \frac{\pi (-r)^3}{3} )$
= $(\pi r^3 - \frac{\pi r^3}{3} ) - (-\pi r^3 + \frac{\pi r^3}{3} ) $
= $(\frac{2 \pi r^3}{3} ) - (\frac{-2\pi r^3}{3} ) $
A = $\frac{4 \pi r^3}{3} $ for a sphere.
Now, let's find the formula for area of the ring outside. This is fairly similar; in order to get the revolution, we will again define a formula for a circle where x is a variable of y and the function passes the variable line test. This is equal to
$y = \sqrt{r^2 - x^2}$. I recommend you google this on function and graph it on desmos for some intuition.
For the sake of your problem, let's set r^2 to 64. We now have $y = \sqrt{64 - x^2}$. Now I STRONGLY recommend you graph that function.
Remember, this is just a semicircle with no negative y values. Therefore, to reach the actual area, we must multiply by two. From the problem you gave me, I assume we would use the shell method:
This video shows the derivation of the formula for the shell method.
Basically, we will be trying to rotate two times the integral from 5 to 8 for our semicircle. This will give us the ring. Note that $V_r$ represents the area of the ring.
$V_r= 2 \pi \int_{5}^{8} x \sqrt{64 - x^2} dx$. Again, for the derivation of this formula, see the link above.
$V_r= 2 \pi \int_{5}^{8} x \sqrt{64 - x^2} dx$
could do this using u sub but I should be heading off to bed
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