# The following are about an infinite region in the 1st quadrant between y=e^-x and the x-axis.

A) Find the area of the region

B)Find the volume of the solid generated by revolving the region about the y-axis

## area = integral ( e^-x) from 0 to infinitiy

= [ - e^-x] from 0 to infinity
= [-1/e^x] from 0 to infinity

now clearly as x ---> ∞ , 1/e^x --> 0
so we have
( 0 - (-1/1)
= 1

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## A) Oh, I see we're dealing with an infinite region. Well, finding the area of an infinite region is like trying to catch a cloud with a butterfly net – it's impossible! The region extends indefinitely, so it doesn't have a finite area. Talk about endless possibilities!

B) Now, let's spin that infinite region around the y-axis. Imagine a never-ending dance party! Unfortunately, calculating the volume of a solid generated by revolving an infinite region is like trying to pour water into a bottomless cup – you'll never fill it up! Since the region extends infinitely, the volume it generates is infinite too. It's like a never-ending flow of surprises! Enjoy the infinite spins!

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## To find the area of the region in the first quadrant between the curve y = e^-x and the x-axis, we can use integration. Here's how you can do it step-by-step:

A) Find the area of the region:

1. Set up the integral: We want to find the area between the curve y = e^-x and the x-axis from x = 0 to infinity. So the integral setup would be:

∫[0,∞] e^-x dx

2. Evaluate the integral: Integrate the function e^-x with respect to x. This can be done using the power rule of integration. The integral of e^-x is -e^-x.

∫[0,∞] e^-x dx = -e^-x evaluated from 0 to ∞

3. Calculate the values: Now, substitute the limits of integration into the expression:

-e^-∞ - (-e^0)

As x approaches infinity, e^-∞ approaches 0, and e^0 is equal to 1.

So, the area of the region is:

-0 - (-1) = 1

Therefore, the area of the region is 1.

B) Find the volume of the solid generated by revolving the region about the y-axis:

To find the volume of the solid generated by revolving the region about the y-axis, we need to use the method of cylindrical shells. Here's how you can do it step-by-step:

1. Set up the integral: We want to find the volume of the solid generated when the region between the curve y = e^-x and the x-axis is revolved about the y-axis from y = 0 to infinity. The integral setup for cylindrical shells is:

V = ∫[0,∞] 2πx f(x) dx

In this case, f(x) is the height of the shell, which is the function y = e^-x. And x represents the radius of the shell.

2. Express f(x) in terms of x: Since we have y = e^-x, we need to solve for x in terms of y to be able to express f(x) in terms of x. Taking the natural logarithm of both sides, we get:

ln(y) = -x

Solving for x, we get:

x = -ln(y)

3. Substitute x and f(x) into the integral: Replace f(x) with e^-x and x with -ln(y) in the integral:

V = ∫[0,∞] 2π(-ln(y)) e^-ln(y) dy

4. Simplify the integral: Use the properties of logarithms to simplify the expression:

V = ∫[0,∞] 2π(-ln(y)) y^(-1) dy

V = ∫[0,∞] -2π ln(y) / y dy

5. Evaluate the integral: Integrate the expression using techniques like integration by parts or substitution. The integral of ln(y) / y can be found using integration by parts.

V = -2π [y ln(y) - y] evaluated from 0 to ∞

6. Calculate the values: Now, substitute the limits of integration into the expression:

-2π [∞ ln(∞) - ∞] - (-2π [0 ln(0) - 0])

As y approaches infinity, ln(∞) approaches infinity, but ln(0) is undefined. However, as y approaches 0, ln(y) approaches negative infinity. So, the limit is:

lim(y→0) y ln(y) = 0

Therefore, the volume of the solid is:

-2π ∞ - (0) = -∞

Since the volume cannot be negative, we take the absolute value:

|V| = ∞

Therefore, the volume of the solid generated by revolving the region about the y-axis is infinity.