Volume of Solids with know CROSS SECTIONS

olicoh
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Homework Statement


In Problems 1-5, let R be the region bounded by y = x^2 + 3, y = 2x, x = 0, and x = 4. Each problem will describe the cross sections of a solid that are perpendicular to the x-axis. Write an integral expression that can be used to find the volume of the solid (do not evaluate).

Problem #1. The cross sections are squares with one edge in R.

Homework Equations





The Attempt at a Solution


A=s^2
A(x)= (x^2+3)^2 + (2x)^2
Int[0,4](x^4+10x^2+9)
 
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olicoh said:

Homework Statement


In Problems 1-5, let R be the region bounded by y = x^2 + 3, y = 2x, x = 0, and x = 4. Each problem will describe the cross sections of a solid that are perpendicular to the x-axis. Write an integral expression that can be used to find the volume of the solid (do not evaluate).

Problem #1. The cross sections are squares with one edge in R.

Homework Equations





The Attempt at a Solution


A=s^2
A(x)= (x^2+3)^2 + (2x)^2
The edge of the square has one end on the line y= 2x and the other on y= x^2+ 3. The length of that side is s= x^2+ 3- 2x= x^2- 2x+ 3. That is what you want to square

Int[0,4](x^4+10x^2+9)
 
HallsofIvy said:
The edge of the square has one end on the line y= 2x and the other on y= x^2+ 3. The length of that side is s= x^2+ 3- 2x= x^2- 2x+ 3. That is what you want to square

I know this might seem like a silly question, but how come you subtracted 2x? Is it because the y=x^2 + 3 function is "on top" of the y=2x when you graph it?
 
Volume of Solids with known cross sections

Homework Statement


In Problems 1-5, let R be the region bounded by y = x^2 + 3, y = 2x, x = 0, and x = 4. Each problem will describe the cross sections of a solid that are perpendicular to the x-axis. Write an integral expression that can be used to find the volume of the solid (do not evaluate).

Problem #1. The cross sections are squares with one edge in R.
Problem #2. The cross sections are circles with the diameter in R.

The Attempt at a Solution


Problem #1: A=s^2
A(x)= (x^2+3-2x)^2
Answer: Int[0,4](x^4-4x^3+10x^2-12x+9)

A user (HallsofIvy) already helped me with this problem. I just want to double check I am answering the question correctly.

Problem #2: A=1/4(pi)d^2
A(x)= 1/4(pi)(x^2+3-2x)^2

Am I going down the right path with this one?
 
Last edited:
Yes:
\int_2^4(x^4- 4x^3+ 10x^2- 19x+ 9)dx
and
\frac{\pi}{4}\int_2^4(x^4- 4x^3+ 10x^2- 19x+ 9)dx

But don't start a new thread when you are still asking the same question.
 
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