Find the volume of the resulting solid

[ProBasket]

A ball of radius 11 has a round hole of radius 8 drilled through its center. Find the volume of the resulting solid

here's what i done:
y = sqrt(11^2-x^2)
y = 8 about x-axis

$$\int pi*(8)^2 - pi*(8-sqrt(11^2-x^2))$$

this is where I am stuck at, no clue if my setup is even correct. What would the bounds be and is my setup correct?

 

[HallsofIvy]

It's not at all clear what you are doing. Where is that "pi*(8)²" from?

It looks like you are confusing a "ball" (sphere) with a circle.

Draw a picture: a circle (the sphere seen from the side) with a rectangle inscribed in it (the cylidrical hole). The center of the sphere is at the middle of the rectangle. The top of the rectangle marks where the hole exits the sphere. If you draw a line from the center of the sphere to the center of the top of the rectangle, it has length h/2 (h is the length of the rectangle). Drawing a line from the center of the sphere to the vertex of the rectangle gives a right triangle with legs of length h/2 and 8 (the radius of the hole) and hypotenuse of length 11 (the radius of the sphere). By the Pythagorean theorem h²/4+ 64= 121 so h²= (57)(4) and h= 2√(57).

Now:

The volume of the sphere, before the hole is drilled, is $\frac{4\pi r^3}{3}= \frac{ (4 \pi(11)^3}{3}$.
The volume of the cylinder removed is $\pi r^2 h= \pi (8)^2(2\sqrt{57})= 128\sqrt{57}\pi$.

BUT you still need to subtract the volume of those two little "caps" that are removed on each end of the cylinder- that's the calculus part of this problem.
Since they are clearly the same, you can calculate the volume of one and double it.
You can do that using the "disk" method. At height "y", The x coordinate is $\sqrt{121- y^2}$ and that becomes the radius of each disk: the area of each such disk is $\pi(121- y^2)$ so the volume is $\pi \int (121- y^2)dy$.
The limits of integration are y= √(57) (the top of the cylinder) to y= 11 (the top of the sphere.

 

[thepatientmental]

Your setup is correct, but there are a few small errors. First, when finding the volume of a solid of revolution, you need to use the formula V = pi * integral of (R(x))^2 dx, where R(x) is the radius of the solid at a given point x. In this case, R(x) would be the distance from the x-axis to the outer edge of the solid, so it would be 11-sqrt(11^2-x^2).

Secondly, when finding the bounds, you need to consider the limits of integration. In this case, the solid is being formed by rotating a semicircle (half of the ball) around the x-axis. So, the limits of integration would be from x = -11 to x = 11, since the semicircle has a radius of 11.

Finally, when setting up the integral, you need to make sure you are using the correct function for the outer radius. In this case, it would be (11-sqrt(11^2-x^2))^2. So, the correct integral would be V = pi * integral from -11 to 11 of (11-sqrt(11^2-x^2))^2 dx.

Solving this integral would give you the volume of the resulting solid. I would recommend using a calculator or a computer program to evaluate the integral, as it can be quite complex to do by hand. I hope this helps clarify any confusion and helps you solve the problem!