Find the volume?

Question

The region in the first quadrant between the x-axis and the graph of y=6x-x^2 is rotated around the y-axis. The volume of the resulting solid of revolution is given by

 

a) ∫ pi(6x-x^2)^2 dx from 0 to 6

 

b) ∫ (2pi*x)(6x-x^2)dx from 0 to 6

 

c) ∫ (pi*x)(6x-x^2)^2 dx from 0 to 6

 

d) ∫ pi(3+sqrt(9-y))^2 dy from 0 to 6

 

e) ∫ pi(3+sqrt(9-y))^2 dy from 0 to 9

 

 

Answer: B

 

Please show all your work.

Answer 1

This is a case of the cylindrical method, since we are rotation around the y axis and integrating over the x axis (because we have a function of x).

 

Visualizing it we have a surface in the x-y plane that we are then going to use like a "radius" for our solid of revolution. So to form our cylinder we have (arc) x (Area), where Area is given by (height) x (length). The solid of revolution is a volume, because the radius is an area, but it is treated as a circumference calculation because the area is "moved through" the circumference. But is must be set up properly in order to work this way.

 

The arc is easy: we are going all the way around so it is 2π.

 

The radius is x is one component of the area, and since it is what we are integrating with respect to it, it  will simply be itself. We also need to determine where our bounds of integration are. We will find them below when we consider the function y.

 

The height of our area is our function y=6x-x². To find our bounds of integration we consider when this function is in the first quadrant. When x is zero, y=0. Recognizing that our parabola opens downward, we determine that if there is a second x-intercept in the first quadrant, it must be the other end of our bound, so we solve    0=6x-x²     ==>     x²=6x     ==>     x=6. So our other bound is x=6.

 

Putting these together we have  (2π) *  ∫ [ (x) * (6x-x^2)  ]  dx   over the bounds [0,6].

Answer 2

Hint:

 

Think of a section of hollow cylindrical pipe enclosing the y-axis.

 

Let the height be y, the radius x, and the wall thickness dx.

 

So the surface area of our pipe is its circumference, 2pi(x), times its height, y.

 

The volume of our infinitesimally thin pipe is its surface area times its thickness, dx.

Answer 3

First you should draw the graph of f(x) in the first quadrant to see what region we are talking about:

since f(x)=6x-x²=-(x-3)²+9, it is the region between the x-axis and an upside down parabola with vertex at (3,9) and x-intercepts 0 and 6. If we wanted the area of this region, we would simply find ∫₀⁶ f(x) dx.

Now imagine this parabola rotating about the y-axis, generating a solid of revolution. To find the volume of this solid you can use the ring method: Think of the solid as being composed of infinitely many rings of radius x, height f(x), and thickness dx. The volume of a thin ring is (circumference)*(height)*(thickness).

Each ring therefore has an infinitesimal volume of

dV = 2pi*x*f(x) dx= 2*pi*x*(6x-x²) dx.

To get the volume of the solid of revolution, you integrate over all rings from 0 to 6:

V= ∫₀⁶ (2pi*x)(6x-x²)dx,

which you can then evaluate.

Answer 4

Hi Sun,

 

There's really virtually no work to show.  I recommend looking up and studying "the Shell Method".  Do this, and let us know what you think/what your thought processes are!