# Thread: Find volume of graph through one revolution on an axis

1. ## Find volume of graph through one revolution on an axis

I have two questions here, the first i need someone to show me my mistake somewhere, the second I have no idea how to do.
1. The area defined by the inequalities
$0\leq y\leq x^2$, $0\leq x\leq 2$
is rotated about the y-axis. Find the volume generated.

I take the raduis of the figure generated from the y-axis outward. so then its value would be $x_{further}-x_{nearer}=2-\sqrt{y}$
$V=\int^4_0 \pi x^2 dy$
$V=\pi \int^4_0(2-\sqrt{y})^2 dy$
$V=\pi \int^4_0(4-4\sqrt{y}+y) dy$
$V=\pi [4y-\frac{8}{3}y^{\frac{3}{2}}+\frac{1}{2}y^2]^4_0$
$V=\frac{8\pi}{3}$
answer is $8\pi$

2. Find the volume generated when the area in the first quadrant enclosed by $y=|x^2-1|$, and the line y=1 is rotated about the line y=1.
I only know that the lines intersect at x=0 and $\sqrt{2}$. Also the radii of the figure would be $1-|x^2-1|$.

Thanks!

2. Originally Posted by arze
I have two questions here, the first i need someone to show me my mistake somewhere, the second I have no idea how to do.
1. The area defined by the inequalities
$0\geq y\geq x^2$, $0\geq x\geq 2$
is rotated about the y-axis. Find the volume generated.

I take the raduis of the figure generated from the y-axis outward. so then its value would be $x_{further}-x_{nearer}=2-\sqrt{y}$
This rotates the region under the curve, you should be rotating the region above it.

$V=\int^4_0 \pi x^2 dy$
$V=\pi \int^4_0(2-\sqrt{y})^2 dy$
$V=\pi \int^4_0(4-4\sqrt{y}+y) dy$
$V=\pi [4y-\frac{8}{3}y^{\frac{3}{2}}+\frac{1}{2}y^2]^4_0$
$V=\frac{8\pi}{3}$
answer is $8\pi$

2. Find the volume generated when the area in the first quadrant enclosed by $y=|x^2-1|$, and the line y=1 is rotated about the line y=1.
I only know that the lines intersect at x=0 and $\sqrt{2}$. Also the radii of the figure would be $1-|x^2-1|$.

Thanks!
2
For x between 0 and 1, $1-|x^2-1|=1-\left(1-x^2\right)$
For x between 1 and $\sqrt{2},\ 1-|x^2-1|=2-x^2$

You can rotate the 2 regions seperately about y=1.

3. Oops I made a mistake in the equations, the signs are the other way around
its
$0\leq y\leq x^2$ and $0\leq x\leq 2$

4. It was wrong for the x,
are you sure the y isn't $x^2\le{y}\le{4}$ ?

5. No its right now its $0\leq y\leq x^2$

6. Sorry arze,

you have to subtract the volume calculated using $r=\sqrt{y}$
from the volume calculated using r=2.

$8\pi$ is correct.

Your volume of rotation using washers is the difference between 2 volumes of rotation.

7. here's the graph

8. sorry i don't get it.
I thought i already calculated the radii, by the difference between the two distances from the y-axis?

9. No, the length you calculated is the width of the washer at any height y.
That's the distance between the pink circles.
As the shape is being revolved around the y axis,
you have 2 radii, one outer radius=2 and one inner radius = $\sqrt{y}$

Hence you must calculate

$\pi\int_{0}^4{2^2}dy-\pi\int_{0}^4{\left(\sqrt{y}\right)^2}dy$