Trouble finding the integral for volume

In summary, The conversation is about finding the integral needed for solving volume problems. The person is asking for help and someone provides hints for the first problem, 56a. They mention that the side of the squares is determined by the difference between two functions, and that integrating over x will give the total volume. The conversation also mentions evaluating the integral for finding the area enclosed by two lines and mentions that the process is easier for problem 56b.
  • #1
Pseudo Statistic
391
6
I'm having trouble finding the integral I'm supposed to use for some Volume problems...
Can someone lead me in the direction as to how I should form my integrals to get the solutions?
The below is a scanned page from an AP Calculus textbook, I'm pretty much stumped on how to solve 56-59..
Hope someone can help.
Thanks.

http://www.brokendream.net/xh4/apcalcscan.jpg
 
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  • #2
I'll give you some hints on the first one (56a). They're all pretty much the same. The side of the squares are determined by the difference between the two functions y=x+1 and y=x^2-1. This difference is x-x^2+2, it is zero for x=-1 and x=2. So now you have determined the shape of your base.

With this you can easily find the area of such a square. Integrating over x gives you the total volume.
 
  • #3
OK, so you merely evaluate the integral [tex]A (x) = \int_{-1}^\2 2 x - x^{2} + 2 dx[/tex]?
Does anybody have a clue about the other questions?
Thanks.
 
Last edited:
  • #4
Well, if you want to evalute the area enclosed by the two lines (y=..) yes, but...
 
  • #5
Pseudo Statistic said:
OK, so you merely evaluate the integral [tex]A (x) = \int_{-1}^\2 2 x - x^{2} + 2 dx[/tex]?
Does anybody have a clue about the other questions?
Thanks.
[itex]x - x^2+2[/itex] gives the length of one side of the square as a function of x. You need the area of the square.

b) Is somewhat easier, since the height of each rectangle is one, that means the area of a cross sectional rectangle is [itex]x-x^2+2[/itex].
 

1. Why is it important to find the integral for volume?

Finding the integral for volume is important because it allows us to calculate the volume of complex three-dimensional shapes, which is useful in many fields of science and engineering.

2. What is the process for finding the integral for volume?

The process for finding the integral for volume involves setting up a triple integral, which represents the volume under the surface of a three-dimensional shape. This triple integral is then solved using calculus techniques.

3. What are common challenges when finding the integral for volume?

One common challenge when finding the integral for volume is setting up the correct limits of integration. This requires a good understanding of the shape and its boundaries. Another challenge is choosing the correct integration method, such as using cylindrical or spherical coordinates.

4. How can I check if my calculated integral for volume is correct?

You can check your calculated integral for volume by using a different method, such as using a different set of coordinates or using a computer program. You can also check your answer by using basic principles of geometry, such as making sure the volume is positive and reasonable for the given shape.

5. Are there any tips for finding the integral for volume more efficiently?

Some tips for finding the integral for volume more efficiently include breaking the shape into smaller, simpler shapes and integrating each part separately. Another tip is to carefully choose the coordinate system and limits of integration to make the integral easier to solve. Additionally, using computer programs or graphs can also help with finding the integral more efficiently.

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