Volume of cross sections using isosceles right triangle

In summary, the area is bounded by the equation y=x^2, x-axis and the line x=3 and is perpendicular to the x axis. The Attempt at a Solution is to find V=x^5/10 from 0 to 3 which is equal to 243/10 .
  • #1
mikaloveskero
6
0

Homework Statement


The area is bounded by the equation y=x^2, x-axis and the line x=3 and is perpendicular to the x axis


Homework Equations


A=1/2bh
Height=Base


The Attempt at a Solution


A=1/2B^2
V=1/2B^2 from 0 to 3
V=1/2 B^3/3 from 0 to 3
V=B^3/6 from 0 to 3
V=27/6

please tell me if I am doing something wrong, on the right track or I'm correct
 
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  • #2
Are we to assume that the cross sections perpendicular to the x-axis are "isosceles right triangles"- with one leg perpendicular to the x-axis or the hypotenuse?

I think what you are trying to do is correct but your notation is very strange. You are given information about x and y- use them, not new variables! Assuming it is the leg that lies perpendicular to the x-axis then the other leg is just y= [itex]x^2[/itex]. The area of such a triangle is [itex](1/2)(x^2)(x^2)= x^4/2[/itex]. The "thickness" of each such cross-section is [itex]\Delta x[/itex] so the volume is [itex](1/2)x^4\Delta x[/itex]. In the limit that becomes an integral.
 
  • #3
HallsofIvy said:
Are we to assume that the cross sections perpendicular to the x-axis are "isosceles right triangles"- with one leg perpendicular to the x-axis or the hypotenuse?

I think what you are trying to do is correct but your notation is very strange. You are given information about x and y- use them, not new variables! Assuming it is the leg that lies perpendicular to the x-axis then the other leg is just y= [itex]x^2[/itex]. The area of such a triangle is [itex](1/2)(x^2)(x^2)= x^4/2[/itex]. The "thickness" of each such cross-section is [itex]\Delta x[/itex] so the volume is [itex](1/2)x^4\Delta x[/itex]. In the limit that becomes an integral.

Its perpendicular to the x-axis not the hypotenuse. Also I combined the variables as one because the height is equal to the base. Also, we weren't taught thickness so that information is not needed. However , in my problem i realize i forgot to plug in X^2

So redoing my work it would be :
V=x^5/10 from 0 to 3 which is equal to 243/10 , therefore your answer should agree with mine after you integrate your work. I'm just not sure where the thickness part come in since i have not learned that yet
 

FAQ: Volume of cross sections using isosceles right triangle

What is the definition of "Volume of cross sections using isosceles right triangle"?

The volume of cross sections using isosceles right triangle refers to the measurement of the space occupied by a three-dimensional shape that is formed by slicing a larger shape using an isosceles right triangle as the cutting tool.

How do you calculate the volume of cross sections using isosceles right triangle?

To calculate the volume of cross sections using isosceles right triangle, you will need to determine the area of the base shape and the height of the resulting shape formed by the isosceles right triangle. Then, you can use the formula V = (1/3) * B * h, where V is the volume, B is the area of the base shape, and h is the height of the resulting shape.

What is the significance of using an isosceles right triangle as the cutting tool for finding the volume of cross sections?

The use of an isosceles right triangle as the cutting tool for finding the volume of cross sections is significant because it allows for consistent and symmetrical cross sections to be formed. This makes it easier to calculate the volume accurately.

What are some real-life applications of finding the volume of cross sections using isosceles right triangle?

Finding the volume of cross sections using isosceles right triangle has various real-life applications, such as in construction and architecture, where it is used to calculate the volume of materials needed for a structure. It is also used in engineering and manufacturing to determine the volume of materials needed for a specific product.

Are there any limitations or assumptions when using this method to calculate volume?

One limitation of using this method to calculate volume is that it assumes the cross sections are uniform and symmetrical. This may not always be the case in real-life situations. Additionally, the shape of the base and the cutting tool must be known in order to accurately calculate the volume.

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