Find Area & Volume of Graph: Region R URGENT!

[needsomehelpinc]

URGENT! Need help with finding Area and Volume of a Graph...

Here is the problem Let R be the region bounded by the graphs of y= square root of x , y= e^-3x and x=1

So I sketch it out and this is what I get for the Area

Integral (0,1) square root of x - Integral (0,1) e^-3x

I took the whole area of the greaph and subtracted it by the small on where i get the anser of .3499

I also need help with finding the volume of the solid generated when region R is revolved about the line y=1, and the same thing when it is revolved around x=2

If some one could help me out that would be great its really urgent

 

[Ryo124]

Area of region R?

1. The problem:

Let R be the region bounded by the graphs of y = √x , y = e^(-3x) , and the line x =1. Find the area of region R.

2. Question:

I'm don't know how to set up the integral for area.. Is it in terms of y or x? HELP PLEASE.

 

[berkeman]

Welcome to the PF, Ryo124. To set up the integral for an area problem like this, you first choose the most convenient coordinates to use, based on the shape and description of the boundaries. Like, if the boundary is circular or arc-based, you might choose circular (2-D) coordinates.

For this one, it looks like a rectangular (x,y) coordinate system is the most natural. Next, draw the three lines specified, and from that, figure out what the region looks like. Then you will set up your <<error text edited out>> integral to make the math the easiest -- either add up a bunch of skinny horizontal dy rectangles or a bunch of skinny vertical dx lines -- whichever makes the math and the limits of the boundaries seem to work the best.

Does that make sense?

 

[Ryo124]

??

Thanks for the welcome. I know how to solve problems like this one, but this particular one is just giving me some problems. I really just need to see how the equation is set up. From there, I can solve.

Can someone please set up the integral for the area of this region??



A = $$\int{\sqrt{x}-e^{-3x}}$$ evaluated from 0 to 1

Something like this maybe?

 

[berkeman]

No, you are not integrating those terms -- those lines define the boundary of the shape that you want to find the are of. Draw the three lines, and see what the x and y limits of the shape are. Then decide which axis you want to integrate over (sorry, I misspoke about a double integral in my previous post -- it's a single integral over one axis or the other). Then the term inside the integral is the height of each skinny rectangle (if you choose to integrate over all the applicable dx on the x axis), and the height of the rectangle is determined by a function that you write which corresponds to the top and bottom boundaries.

Maybe start with a simpler example problem first. Define the area as lying between the lines y=x, x=1, and y=0 (a triangle in the first quadrant). Then set up the integral to calculate its area:

$$A = \int { (what?)} dx$$

where the limits of integration are x = 0,1 (I don't know how to put the limits on the LaTex integral sign yet).

 

[Ryo124]

Then the term inside the integral is the height of each skinny rectangle (if you choose to integrate over all the applicable dx on the x axis), and the height of the rectangle is determined by a function that you write which corresponds to the top and bottom boundaries.

The top boundary is y = $$\sqrt{x}$$ and the bottom boundary is y = $$e^{-3x}$$

Therefore, wouldn't the height be $$\sqrt{x}-e^{-3x}$$ ?

Sorry if I'm being stubborn, I just can't set up the equation correctly and I'm having problems deciding what the limits are. If you look at the graph it would be easier to understand.

 

[berkeman]

Oh, if those are the y-bounds of the area, then yes, that would be the height that you would want to integrate over the x interval. Sorry for being dense.

To find the x interval, you'll need to set the two terms equal to see where they cross on the left and right of the area.

 

[Gib Z]

You must first think, do $y=\sqrt{x}$ really intersect $y=e^{-3x}$ at zero? Use what you know about logs to solve that.

So the bounds on your integral is wrong, your area is wrong.

Get that figured out first before you move onto the volume, cause I gtg :P

 

[HallsofIvy]

This was also posted under "homework" so I am combining the two.

 

[Ryo124]

Oh well.. I took the test and most likely failed :yuck: But that's okay. I asked my teacher for help and she showed me how to do it. Thanks anyway for the help guys. I'm sure she'll give me some points for my efforts