Calculating volumes by shell and slicing

[joe007]

Homework Statement

The area under the graph of the function y = cos inverse x on the interval [0; 1] is rotated
about the x-axis to form a solid of revolution.
(a) Write down the volume V of the solid as a denite integral with respect
to x according to the disc/slicing method. Do NOT attempt to evaluate this
integral.
(b) Write down the volume V of the solid as a denite integral with
respect to y according to the shell method.
(c) Using the antiderivative,
Integral y cos y dy = y sin y + cos y + C;
or otherwise, find the volume V of the solid as an exact real number.

Homework Equations

V=2*PI integral x*f(x) dx

The Attempt at a Solution

well

(a)V=integral 1 to 0 Pi*cos^-2x^2 dx

(b)V=2PI integral 1 to 0 ycosy dy

(c) i solved it and i got 2PI(PI/2 -1)

 

[dynamicsolo]

Your function is y = \cos^{-1} x . Be aware that this is inverse functional notation, so \cos^{-1} x \neq \frac{1}{\cos x} . So when you write the integral for the "disc/washer method", it should be

V = \pi \int_{0}^{1}(\cos^{-1}x)^{2} dx ,

since you are squaring an inverse cosine function.

On part (b), watch out when working with y as the variable of integration. What does the graph of inverse cosine x from 0 ≤ x ≤ 1 look like? When you "turn it sideways" to make the slices parallel to the x-axis for the shells, is there only one set of boundaries for the "lower" and "upper ends" of the shells?

 

[joe007]

the y values should be from 0 to PI/2 therefore how would i take a slice of the shell, would i just integrate ycosy dy :P

 

[dynamicsolo]

You really should make a graph of the region that is being revolved about the x-axis: the curve x = cos y is only a boundary for part of it if you integrate in the y-direction (as you will have to for shells).

What ARE the boundaries of the region and what will act as the "upper" and "lower curves" if you are integrating with respect to y? (Using y cos y is only a part of this.)

 

[dynamicsolo]

(Forget that last note: now that I look at this again this morning, I see that the grapher I used took y = cos^-1 x and plotted y = sec x anyway. )

The curve does close properly from (0, pi/2) to (1, 0). You will need to change your limits of integration.

So using shells, you should have 2\pi \int_{0}^{\pi/2} y cos y dy .

Sorry for my earlier comment.

Your answer for part (c) appears to be correct. So you integrated correctly, but wrote the limits wrong in part (b).

 

[k990610]

Hi, I was just passing by and I don't follow this integrations..
Sorry but dynamicsolo can you please explain in details how to get to the answers for (b) and (c)?

 

[joe007]

hmmm its a method of integration to calculate volumes by the shell method

 

[dynamicsolo]

hmmm its a method of integration to calculate volumes by the shell method

He's asking exactly how you set up the shells; we're discussing this by PM.