Volume by parallel cross-section.

[Call my name]

The questions: the base of a solid is the region bounded y=x^2 and y=4. Find the volume of the solid given that the cross sections perpendicular to the x-axis are squares.

The answer is 512/15.

I set A(x)= (2y)^2 = 4(x^2)^2=4x^4, and the answer is wrong when I integrate this from x = -2 to x = 2.

How can I solve this?

Please help me.

 

[Mark44]

The typical volume element, \Delta V, has a volume of (4 - x²)(4 - x²)\Delta x. The first factor of (4 - x²) is the distance between the line y = 4 and the curve y = x². The second factor of (4 - x²) is the vertical height of the volume element (perpendicular to the x-y plane). The \Delta x factor is the thickness of the volume element.

 

[Call my name]

That's more confusing... can you explain it with integrals?

 

[LCKurtz]

Maybe this picture will help you:

 

[Call my name]

Maybe this picture will help you:

Thank you for the picture, but I do understand how it looks...

just don't know how it goes with the integrals...

 

[Call my name]

The typical volume element, \Delta V, has a volume of (4 - x²)(4 - x²)\Delta x. The first factor of (4 - x²) is the distance between the line y = 4 and the curve y = x². The second factor of (4 - x²) is the vertical height of the volume element (perpendicular to the x-y plane). The \Delta x factor is the thickness of the volume element.

so A(x) = (4-x²)² ?
2\int(4-x²)²dx (x is 0~2)

 

[LCKurtz]

Yes, since you have symmetry about x = 0.

 

[Call my name]

Yes, since you have symmetry about x = 0.

But, I am not getting 512/15.

 

[Mark44]

Then you have done something wrong.
\int_{x = -2}^2 (4 - x^2)^2 dx~=~2\int_{x = 0}^2 (4 - x^2)^2 dx~=~\frac{512}{15}

 

[Call my name]

Yes. made some calculation error

 

[Mark44]

Did you find out where you went wrong?

 

[Call my name]

Did you find out where you went wrong?

Yes. Thank you. I have posted another question this time regarding shell method