R is the region bounded above by the graph of f(x)=2e^−x^2 and below by the x-axis over the interval [1,2]. Find the volume of the solid of revolution formed by revolving R around the y-axis.

Question

R is the region bounded above by the graph of f(x)=2e^−x^2 and below by the x-axis over the interval [1,2]. Find the volume of the solid of revolution formed by revolving R around the y-axis.

Answer 1

using shells of thickness dx,

v = ∫[1,2] 2πrh dx
where r = x and h = y
v = ∫[1,2] 2πx*2e^(-x^2) dx = 2π(e^3-1)*e^-4
do this by letting u = x^2 and you have ∫[1,4] 2πe^-u du

Answer 2

or, using discs (washers) of thickness dy,

v = ∫[2/e^4,2/e] π(R^2-r^2) dy
where r = 1 and R^2 = ln(2/y)
v = ∫[2/e^4,2/e] π*(ln(2/y) - 1) dy = 2π(e^3-1)*e^-4

Answer 3

To find the volume of the solid of revolution formed by revolving the region R around the y-axis, we can use the method of cylindrical shells. The formula for the volume of a solid of revolution using cylindrical shells is:

V = ∫(2πxf(x))dx

In this case, we need to express the function f(x) in terms of y, as we are revolving the region around the y-axis. So, let's solve for x in terms of y:

f(x) = 2e^(-x^2)
y = 2e^(-x^2)

Now, we need to find x in terms of y. Taking the natural logarithm of both sides, we get:

ln(y/2) = -x^2

Solving for x, we have:

x = ±√(-ln(y/2))

Since we are revolving the region R around the y-axis, the radius x becomes the height of the cylindrical shell. Therefore, the height of the cylindrical shell is:

h = x = ±√(-ln(y/2))

Now, let's find the limits of integration. Since R is bounded between the x-axis and the graph of f(x), we need to find the y-values corresponding to x = 1 and x = 2.

For x = 1:

y = 2e^(-1^2) = 2e^(-1) = 2e^(-1)

For x = 2:

y = 2e^(-2^2) = 2e^(-4) = 2e^(-4)

So, the limits of integration for y are [2e^(-4), 2e^(-1)].

Now, we can calculate the volume using the formula:

V = ∫(2πxf(x))dx
= ∫(2πy)dx
= 2π∫(±√(-ln(y/2)))dy

We integrate with respect to y since we expressed x in terms of y. The ± sign indicates that we need to consider both positive and negative values of x.

Let's evaluate this integral using the given limits of integration [2e^(-4), 2e^(-1)]:

V = 2π∫(±√(-ln(y/2)))dy, where y ranges from 2e^(-4) to 2e^(-1)

Now, you can evaluate this integral using appropriate techniques like u-substitution, or numerical methods like approximating the integral using numerical integration techniques.

Once you have the definite integral evaluated, you will get the volume of the solid of revolution formed by revolving R around the y-axis.