I'm having trouble with the following problem:

Question

I'm having trouble with the following problem:

Find the volume of the solid generated by revolving the region about the given line. The region is in the first quadrant bounded above by the line y= sqrt 2, below by the curve y=secxtanx, and on the left by the y-axis. Rotate the region about the line y=sqrt 2.

The problem is supposed to be done using the disk method.
So far I know that I need to find the point of intersection betwen y= sqrt 2 and y=secxtanx
secxtanx= sqrt 2
I'm not sure how to do the math to find that, but when I used my calculator to approximate the intersection I got .78539816. I know that this will be my upper bound, and the lower bound is zero. I also know the general formula for volume is
v= pi (integral from a to b) [f(x)^2]dx
I also know that the answer is supposed to be pi(pi/2 + 2sqrt2 + 11/3). However, I don't know how to actually solve the problem.

Answer 1

To find the point of intersection between the line y = sqrt 2 and the curve y = secxtanx, we can set them equal to each other:

sqrt 2 = secxtanx

Taking the reciprocal of both sides, we get:

1/sqrt 2 = cscxtanx

Simplifying further:

sqrt 2 / 2 = sinx/cosx

Dividing both sides by sqrt 2, we get:

1 / sqrt 2 = sinx / (cosx * sqrt 2)

Applying the identity sin(45 degrees) = 1 / sqrt 2, we find:

sin(45 degrees) = sinx / (cosx * sqrt 2)

Since the sine function is positive in the first quadrant, we can conclude that x = 45 degrees.

Now, let's proceed with finding the volume using the Disk Method.

First, let's set up the integral for the volume:

V = pi * (integral from a to b) [f(x)^2]dx

In this case, the lower bound a is 0, and the upper bound b is x = 45 degrees (or pi/4 in radians).

The function f(x) represents the distance from the line y = sqrt 2 to the curve y = secxtanx. We can express f(x) as:

f(x) = sqrt 2 - secxtanx

Now, let's substitute the values in the integral:

V = pi * (integral from 0 to pi/4) [sqrt 2 - secxtanx]^2 dx

Expanding the square, we get:

V = pi * integral from 0 to pi/4 [2 - 2sqrt(2) secxtanx + sec^2xtan^2x] dx

Using trigonometric identities, we can simplify further:

V = pi * integral from 0 to pi/4 [2 - 2sqrt(2) secxtanx + sec^2xtan^2x] dx

V = pi * [2x - 2sqrt(2) ln|secxtanx + tanx| + tanx + C] evaluated from 0 to pi/4

Removing the absolute value, we have:

V = pi * [2(pi/4) - 2sqrt(2) ln|sec(pi/4)tan(pi/4) + tan(pi/4)| + tan(pi/4)] - [2(0) - 2sqrt(2) ln|sec(0)tan(0) + tan(0)| + tan(0)]

Simplifying further:

V = pi * [pi/2 - sqrt(2) ln(2 + 1) + 1] - [0 - 2sqrt(2) ln(1) + 0]

V = pi * [pi/2 - sqrt(2) ln(3) + 1] - [0 - 0 + 0]

V = pi * (pi/2 - sqrt(2) ln(3) + 1)

Finally, the answer for the volume is pi * (pi/2 + 2sqrt(2) + 11/3).

I hope this helps! Let me know if you have any further questions.

Answer 2

To solve the problem, we first need to find the point of intersection between the two functions y = √2 and y = sec(x)tan(x). Let's set them equal to each other and solve for x:

√2 = sec(x)tan(x)

Now, to find the intersection point precisely, we can use trigonometric identities to simplify the equation. For example, we know that tan(x) = sin(x)/cos(x) and sec(x) = 1/cos(x). Substituting these identities, the equation becomes:

√2 = (1/cos(x))(sin(x)/cos(x))

Multiplying both sides by cos(x), we get:

√2 cos(x) = sin(x)

Now, let's square both sides to eliminate the square root:

2 cos^2(x) = sin^2(x)

Using the Pythagorean identity, sin^2(x) + cos^2(x) = 1, we can rewrite the equation as:

2 cos^2(x) = 1 - cos^2(x)

Rearranging the equation:

3 cos^2(x) = 1

Dividing both sides by 3:

cos^2(x) = 1/3

Taking the square root:

cos(x) = ±√(1/3)

Now we can find the value of x by taking the inverse cosine:

x = cos^(-1)(±√(1/3))

Using a calculator to evaluate this expression, we find that x is approximately ±0.9553 radians.

Since we are only interested in the value in the first quadrant, we take the positive value:

x ≈ 0.9553 radians

Now that we have the bounds, we can calculate the volume using the disk method. The integral setup will be:

V = π ∫[0 to 0.9553] [f(x)^2] dx

The function f(x) represents the distance between the line of revolution (y = √2) and the curve (y = sec(x)tan(x)). In this case, it is given by f(x) = √2 - sec(x)tan(x).

So the final integral becomes:

V = π ∫[0 to 0.9553] [(√2 - sec(x)tan(x))^2] dx

Evaluating this integral will give you the volume of the solid generated by revolving the region about the line y = √2.

Answer 3

Yes, you have done your homework, and you are not far from finishing it.

First, the disk method is well described in the following article. In the later part of the solution, if you have problems, you can refer to it.
http://www.vias.org/calculus/06_applications_of_the_integral_02_02.html

A sketch of the volume to be found can be found at the following link:
http://img85.imageshack.us/img85/1486/jenna.png

Your expression for finding the volume by the disk method is correct. All that is missing is the function for the "radius" of the elemental disks, and the limits a and b.

We know from the question description that the lower limit is 0. The upper limit is as you have found it, π/4.
It can be found as follows:
tan(x)/cos(x) = sqrt(2)
By expanding tan(x) into sin/cos, and applying the identity sin²+cos&sup2 = 1 you will end up with a quadratic equation in sin(x),
sqrt(2)sin²(x)+sin(x)-sqrt(2)=0
of which the positive solution is x=π/4.

The function f(x) representing the radius of the solid of rotation expressed as a function of x is evident from the sketch:
f(x) = sqrt(2)-tan(x)/cos(x)
So proceed with the definite integral and you will succeed in finding the correct answer.