Expert: Paul Klarreich - 5/13/2007

Question
let R be the region enclosed by the graphs of by=cost(x^2) and by=x^2.
a). set up and evaluate an integral that gives the area of R.
B).without using absolute value, set up an expression involving one or more integrals that gives the volume of the solid generated by revolving R about the x-axis.do not evaluate.
ac).without using absolute value, set up an expression involving one or more integrals that gives the volume of the solid generated by revolving R about the line x=1.do not evaluate.
thanks

Answer
Questioner:   bhavika
Category:  Calculus
 
Subject:  calculus
Question:  let R be the region enclosed by the graphs of by=cost(x^2) and by=x^2.
a). set up and evaluate an integral that gives the area of R.
B).without using absolute value, set up an expression involving one or more integrals that gives the volume of the solid generated by revolving R about the x-axis.do not evaluate.
ac).without using absolute value, set up an expression involving one or more integrals that gives the volume of the solid generated by revolving R about the line x=1.do not evaluate.
thanks
 

let R be the region enclosed by the graphs of by=cost(x^2) and by=x^2.

Did you mean to write:  y = cos(x^2)  and  y = x^2?  What you wrote did not make sense.  PLEASE proofread your question carefully before you hit the 'send' button, because I might misinterpret your question, and then you don't get a good answer.


Assuming that's true, you need the points of intersection of the graphs.  Those are the solutions to the equation:

cos(x^2) = x^2

Unfortunately, there is no exact way to determine these values, which I will call +-A, so all you can do here is write:

{A
|   (cos(x^2) - x^2) dx
}-A

as the area.

If you revolve that about the x-axis, you can integrate by disks.  Each disk has volume  V = pi r^2 h, since it's a cylinder.  Your volume element is:

dV = pi (cos(x^2) - x^2)^2 dx

and you integrate that from  -A to A as before.
.................................
If you revolve it about  x = 1, note that  A < 1, so your area is totally to the left of that line.  Now use cylindrical shells:

V = 2pi rh dx

In this case, your  r is the distance from the shell to x = 1, i.e.  1 - x.
And your h is the function.

dV = 2pi (cos(x^2) - x^2)(1 - x) dx

and you integrate that from  -A to A as before.