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?"

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reply Mar 7 "14 in ~ 4:17

DJohnMDJohnM
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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

G Tony JacobsG Tony Jacobs
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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

http://symbolab.com/solver/step_by_step/2%5Cpi%5Cint%5E%7B8%7D_%7B5%7Dx%5Csqrt%7B64%20-%20x%5E%7B2%7D%7Ddx%20