# Math Help - Volume by Slicing

1. ## Volume by Slicing

The solid lies between planes perpendicular to the x axis at $x = 0$ and $x = 4$. The cross-sections perpendicular to the x axis on the interval $0 \leq x \leq 4$ are squares whose diagonals run from the parabola $y = -\sqrt{x}$ to the parabola $y = \sqrt{x}$. Find the volume
First step?

$V = \int_{a}^{b} A(x) dx$

$x^{2}$ is the area of a square

$A = x^{2}$

$V = \int_{0}^{4} [2 \sqrt{x}]^{2} dx$

$V = \int_{0}^{4} 4x dx$

$V = \dfrac{4x^{2}}{2}$

$V = 2x^{2}$

$V = [2(4)^{2}] - [2(0)^{2}]$

$V = [32] - [0] = 32$ The book says the answer is 16. Something went wrong here.

2. ## Re: Volume by Slicing

The length of the diagonal of the square is $2\sqrt{x}$, so the length of a side is $\frac{1}{\sqrt{2}}$ times $2\sqrt{x}$, which is $\sqrt{2x}$. So the area is 2x, not 4x.

- Hollywood

3. ## Re: Volume by Slicing

How could we visualize this?

4. ## Re: Volume by Slicing

The square is rotated so that the diagonal is in the x-y plane, so the sides of the square are 45 degrees from horizontal. I imagine it starting as a point (x=0) and growing as it comes toward me (x increasing) until it reaches x=4, where the vertices are (4,2,0), (4,0,2), (4,-2,0), and (4,0,-2).

Is that what you meant by "How could we visualize this?"

- Hollywood

5. ## Re: Volume by Slicing

The solid lies between planes perpendicular to the x axis at $x = 0$ and $x = 4$. The cross-sections perpendicular to the x axis on the interval $0 \leq x \leq 4$ are squares whose diagonals run from the parabola $y = -\sqrt{x}$ to the parabola $y = \sqrt{x}$. Find the volume
$V = \int_{a}^{b} A(x) dx$

$A = [\dfrac{2 \sqrt{x}}{2}]^{2}$

$A = \dfrac{4x}{4}$

$A = x$

$V = \int_{0}^{4} x dx$

$V = \dfrac{x^{2}}{2}$

$V = \dfrac{(4)^{2}}{2} - 0$

$V = \dfrac{16}{2} = 8$ Again, answer in book is 16. Where am I going wrong here?

How about this way?

$V = \int_{a}^{b} A(x) dx$

$A = [\dfrac{2 \sqrt{x}}{2}]$

$A = \dfrac{2\sqrt{x}}{2}$

$A = \sqrt{x}$

$V = \int_{0}^{4} \sqrt{x} dx$

$V = \int_{0}^{4} x^{1/2} dx$

$V = \dfrac{x^{3/2}}{\dfrac{3}{2}}$

$V = \dfrac{2}{3}x^{3/2}$

$V = \dfrac{2}{3}(4)^{3/2} - 0 = \dfrac{16}{3}$ Also wrong

Finally, how about this way?

The solid lies between planes perpendicular to the x axis at $x = 0$ and $x = 4$. The cross-sections perpendicular to the x axis on the interval $0 \leq x \leq 4$ are squares whose diagonals run from the parabola $y = -\sqrt{x}$ to the parabola $y = \sqrt{x}$. Find the volume
$V = \int_{a}^{b} A(x) dx$

$x^{2}$ is the area of a square

$A = x^{2}$

$V = \int_{0}^{4}[x] [2 \sqrt{x}]^{2} dx$

$V = \int_{0}^{4} [x][4x] dx$

$V = \dfrac{4x^{2}}{2}$

$V = 2x^{2}$

$V = 2(4)^{2} - 0 = 32$

OK, this way looks better. But still not 16. Maybe the diagonal thing needs to be put in.