# Thread: Solids of revolution question

1. ## Solids of revolution question

Can you help me to solve these 2 questions? They are very confusing and I cannot find the right way.
I have the answer from the book but I need the way to solve it.
Thanks a lot!!!!
1) A cylindrical hole of diameter 2 cm is bored through a sphere of radius 4 cm. The hole has been bored so that its main axis passes through the centre of the sphere. Calculate the volume of the remainder of the sphere.
2) Consider that part of the parabola y = (x-1)*(x+2) which lies below the x-axis. Calculate the volume of the solid formed when this part is rotated about the line y =4.

2. ## Re: Solids of revolution question

Originally Posted by tosucceed999
Can you help me to solve these 2 questions? They are very confusing and I cannot find the right way.
I have the answer from the book but I need the way to solve it.
Thanks a lot!!!!
1) A cylindrical hole of diameter 2 cm is bored through a sphere of radius 4 cm. The hole has been bored so that its main axis passes through the centre of the sphere. Calculate the volume of the remainder of the sphere.
The simplest way to do this is to first find the volume of the entire sphere, $(4/3)\pi(4)^3$, then subtract the volume removed. That will be a cylinder plus two "caps" at either end. The radius of the cylinder is, of course, 2 cm. To find the height, h, draw a circle with a thin rectangle inscribed in it. The line from the center of the circle to a corner of the rectangle has length the radius of the sphere, 4 cm. A line from the center of the sphere perpendicular to the side of the rectangle has length the radius of the cylinder, 2 cm. Use the Pythagorean theorem to find half the length of the cylinder. Two find the volume of a cap, think of it as a stack of disks. Setting up a coordinate system with the origin at the center of the sphere, the equation of the sphere's projection in the xy-plane is $x^2+ y^2= 16$ or [itex]x= \pm\sqrt{16- y^2}[/tex] and that is the radius of one of the disks. Its area is $\pi x^2= \pi(16- y^2)$. If its thickness is dx, the volume of one disk is $\pi(16- y^2)dy$ and so the volume of them all is [tex]\pi \int 16- y^2 dy[tex]. The integral is take from the top of the cylinder to the top of the sphere.

2) Consider that part of the parabola y = (x-1)*(x+2) which lies below the x-axis. Calculate the volume of the solid formed when this part is rotated about the line y =4.
I would do this in two parts.

First look at the parabola only. Imagine a line from $y= x^2+ x- 2$ to y= 4. Rotating around the axis y= 4 gives a disk of radius $4- (x^2+ x- 2)$ and so of area $\pi (6- x^2- x)^2$. Taking each disk to have "thickness" dx, The volume of each disk is $\pi (6- x^2- x)^2 dx$ and the volume of all put together is $\pi\int (6-x^2-x)^2 dx$. The integral is, of course, from x= -2 to x= 1 where the parabola crosses the x-axis.

Second, subtract the volume of the cylinder made up of the portion that is above the x-axis. That will be a cylinder with radius 4 and height 1-(-2)= 3.

Thank you.