A cylindrical drill through radius 5 is provided to bore a hole with the facility of a round of radius 8. Discover the volume of the ring shaped solid the remains.

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Alright, my point is that i did no understand just how to set the integral


$egingroup$ i don't think you require an integral. Since the feet is drilled through, the height of the cylinder is $geq8$. So, the area that the sphere is $$frac43picdot8^3$$ The area the the cylinder is $$geqpicdot5cdot8$$ The volume continuing to be is because of this the volume of the round minus the room of the cylinder. $endgroup$
If you Google on "hole with a sphere" or "napkin ring formula" friend will find that the volume that remains as soon as a ball is pierced through a cylinder is dependency only top top the height of the "ring".

In this case, the elevation of the continuing to be ring is $H$:$$H=2 imes (8^2-5^2)$$Treat $fracH2$ together the radius the a sphere, and also the volume that the sphere is the answer.

Some time back there was a loooong conversation in a barisalcity.orgematics group around this problem: " ns drill a 5" lengthy hole centrally v a sphere. What is the volume of the continuing to be ring?"

reply Mar 7 "14 in ~ 4:17

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A round of radius $8$ is derived by revolving a circle of radius $8$ around the $x$-axis. If you swing the an ar between that circle and the heat $y=5$ around the $x$-axis, you acquire the an ar you"re interested in. That means you want to usage the washer method, with $f(x)=sqrt64-x^2$ and also $g(x)=5$.

The borders of integration space the $x$-values of the intersection clues of the graphs that $f$ and also $g$.

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reply Mar 7 "14 in ~ 2:52

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First we should remember just how we derive area the a sphere. We start out with $x^2 + y^2 = r^2$, or the equation the a circle. Dividing both sides by $r^2$ gives us $fracx^2r^2 + fracy^2r^2 = 1$. Individually both sides by $fracx^2r^2$ provides us $fracy^2r^2 = 1 - fracx^2r^2$. By acquisition the square root of both political parties we have $fracyr = sqrt1 - fracx^2r^2$ offering us a final equation the $y = rsqrt1 - fracx^2r^2$. The reason this is necessary is due to the fact that we now have actually an equation for y the is dependent top top x. If you go to desmos and also graph this equation, friend will obtain a half circle through no negative y values. This have to be clear once looking in ~ the equation.

By utilizing the key method, we identify 1. That y is the radius of each disk and also 2. That, since $y = rsqrt1 - fracx^2r^2$, the area of each disk is equal to $pi(rsqrt1 - fracx^2r^2)^2$. Thus each disk has a volume of $pi(sqrtr^2sqrt1 - fracx^2r^2)$ = $pi(sqrtr^2 - x^2)^2$ = $pi r^2 - pi x^2$. Integrating, we have actually

$int_r^-r(pi r^2 - pi x^2 ) dx$

$(pi r^2 x - fracpi x^33 ) $ in between -r and also r

substituting in $r$ and also $-r$ because that b and a, us have

$ (pi r^2 r - fracpi (r)^33 )-(pi r^2 (-r) - fracpi (-r)^33 )$

= $(pi r^3 - fracpi r^33 ) - (-pi r^3 + fracpi r^33 ) $

= $(frac2 pi r^33 ) - (frac-2pi r^33 ) $

A = $frac4 pi r^33 $ for a sphere.

Now, let"s find the formula because that area of the ring outside. This is reasonably similar; in order to acquire the revolution, we will certainly again specify a formula for a circle where x is a variable of y and the role passes the variable heat test. This is same to

$y = sqrtr^2 - x^2$. Ns recommend girlfriend google this on function and graph the on desmos for some intuition.

For the services of your problem, let"s set r^2 to 64. We now have actually $y = sqrt64 - x^2$. Now I strong recommend you graph that function.

Remember, this is simply a semicircle through no negative y values. Therefore, to reach the actual area, we should multiply by two. From the trouble you provided me, i assume us would usage the covering method: https://www.khanacademy.org/barisalcity.org/calculus/solid_revolution_topic/solid_of_revolution/v/solid-of-revolution--part-5

This video clip shows the derivation of the formula because that the covering method.

Basically, we will certainly be do the efforts to turn two time the integral indigenous 5 to 8 because that our semicircle. This will provide us the ring. Note that $V_r$ represents the area that the ring.

$V_r= 2 pi int_5^8 x sqrt64 - x^2 dx$. Again, for the source of this formula, see the attach above.

$V_r= 2 pi int_5^8 x sqrt64 - x^2 dx$

could perform this making use of u sub however I have to be heading off to bed