# Volume by parallel cross-section.

1. Oct 30, 2009

### 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.

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?

Last edited: Oct 30, 2009
2. Oct 30, 2009

### Staff: Mentor

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.

3. Oct 30, 2009

### Call my name

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

4. Oct 30, 2009

### LCKurtz

5. Oct 30, 2009

### Call my name

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

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

6. Oct 30, 2009

### Call my name

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

7. Oct 30, 2009

### LCKurtz

Yes, since you have symmetry about x = 0.

8. Oct 31, 2009

### Call my name

But, I am not getting 512/15.

9. Oct 31, 2009

### Staff: Mentor

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

10. Oct 31, 2009

### Call my name

11. Oct 31, 2009

### Staff: Mentor

Did you find out where you went wrong?

12. Oct 31, 2009

### Call my name

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