Cross-sectional solid: region bdd by y=e^x, x-axis, x=0, x=1

[thatguy47]

Please help me with this question:
Consider the region bounded by y=e^x , the x-axis and the lines x=0 and x=1. Find the volume of the following solids.
#1) The solid whose base is the given region and whose cross-sections perpendicular to the x-axis are squares.
I got the answer on my calculator but how do you do it by hand?
The answer by hand is (e^2 - 1)/2 <from answer key> which is what I get when I solve with the calculator ex. [3.1945... = (e^2 - 1)/2 ]
Here's what I did for the calculator:
side = 0 - e^x
fnint((-e^x)^2,x,0,1) to get 3.1945....
How would you solve this by hand instead of calculator???

 

[skeeter]

Re: Cross-sectional solid question

\(\displaystyle V = \int_0^1 (e^x)^2 dx\)

\(\displaystyle V = \int_0^1 e^{2x} dx\)

\(\displaystyle V = \left[\frac{1}{2} e^{2x} \right]_0^1\)

\(\displaystyle V = \frac{1}{2} \left[e^2 - e^0\right]\)

\(\displaystyle V = \frac{e^2 - 1}{2}\)

 

[thatguy47]

Re: Cross-sectional solid question

Thanks for the reply. Also, how do you draw the integral images in your posts?

 

[galactus]

That's called LaTex. Click on 'quote' at the upper right corner of skeeters post to see how he typed the code.

 

[thatguy47]

Thanks. I have one more question. How would you set up the problem for:
Consider the region bounded by y=e^x , the x-axis and the lines x=0 and x=1. Find the volume of the following solids.
#2) The solid whose base is the given region and whose cross-sections perpendicular to the x-axis are semicircles.

 

[skeeter]

the vertical distance from \(\displaystyle y = e^x\) to the x-axis would be the diameter of each semicircular cross section, so \(\displaystyle d = 2r = e^x\)

area of a semicircle ...

\(\displaystyle A = \frac{\pi r^2}{2}\)

general volume formula for any cross section perpendicular to the x-axis ...

\(\displaystyle V = \int_a^b A(x) \, dx\)



btw ... next time, start a new thread for a new problem.

 

[thatguy47]

the vertical distance from \(\displaystyle y = e^x\) to the x-axis would be the diameter of each semicircular cross section, so \(\displaystyle d = 2r = e^x\)

area of a semicircle ...

\(\displaystyle A = \frac{\pi r^2}{2}\)

general volume formula for any cross section perpendicular to the x-axis ...

\(\displaystyle V = \int_a^b A(x) \, dx\)



btw ... next time, start a new thread for a new problem.

I still don't get how to do it . I know the answer is [(e^2-1)/16]*pi
On the calculator I was doing: fnint((pi/2)((e^x)/2)^2, x, 0, 1) but I'm still not getting the right answer :? .

 

[galactus]

Graph \(\displaystyle e^{x}\). Now, picture a whole bunch of semicircles stacked up along it. Since \(\displaystyle y=e^{x}\), then that is the base of your whole bunch of

semicircles. In other words, the diameter of the semicircles. See?.

As Skeeter pointed out.

The radius of the semicircle is half that: \(\displaystyle \frac{e^{x}}{2}\)

Since the area of a semicircle is \(\displaystyle \frac{1}{2}{\pi}y^{2}\)

We get:

\(\displaystyle \frac{1}{2}{\pi}\int_{0}^{1}\left(\frac{e^{x}}{2}\right)^{2}\)

It's the same idea as the square. Only here we use the area of a semicircle and plug in our function. With the square, since the base of the square was

\(\displaystyle e^{x}\), we squared it to get the area. I hope I described that OK so you understand what I mean.

 

[thatguy47]

Graph \(\displaystyle e^{x}\). Now, picture a whole bunch of semicircles stacked up along it. Since \(\displaystyle y=e^{x}\), then that is the base of your whole bunch of

semicircles. In other words, the diameter of the semicircles. See?.

As Skeeter pointed out.

The radius of the semicircle is half that: \(\displaystyle \frac{e^{x}}{2}\)

Since the area of a semicircle is \(\displaystyle \frac{1}{2}{\pi}y^{2}\)

We get:

\(\displaystyle \frac{1}{2}{\pi}\int_{0}^{1}\left(\frac{e^{x}}{2}\right)^{2}\)

It's the same idea as the square. Only here we use the area of a semicircle and plug in our function. With the square, since the base of the square was

\(\displaystyle e^{x}\), we squared it to get the area. I hope I described that OK so you understand what I mean.

Thanks for explaining it. I totally understand it now and I finished the rest of my problems correctly. Your awesome !