# Volume by integration problem

#### Bmanmcfly

##### Member
[solved]Volume by integration problem

I was just curious if I'm doing this wrong, but for the problem I am working on, I am rotating around the y axis a formula.

Part of the graph crosses into the negative x-axis.

So, my questions are; do I have to separate this into 2 integrals, one for the positive section and one for the negative section like is done with areas?

Also, if I do this split an there is a multiplier, do I keep this multiplier (in this case pi) on both halves? Or would that unintentionally create a 2pi?

Last edited:

#### MarkFL

Staff member
Since you cite $\pi$ as the multiplier, I assume you are using the disk method, and presumably the $x$-coordinate is the radius of the disks, which is squared, so you most likely may use one integral over the entire interval.

In the case of a split, you do want to apply the multiplier to both integrals, since:

$$\displaystyle k\int_a^b f(x)\,dx+k\int_b^c g(x)\,dx=k\left(\int_a^b f(x)\,dx+\int_b^c g(x)\,dx \right)$$

#### Bmanmcfly

##### Member
Since you cite $\pi$ as the multiplier, I assume you are using the disk method, and presumably the $x$-coordinate is the radius of the disks, which is squared, so you most likely may use one integral over the entire interval.

In the case of a split, you do want to apply the multiplier to both integrals, since:

$$\displaystyle k\int_a^b f(x)\,dx+k\int_b^c g(x)\,dx=k\left(\int_a^b f(x)\,dx+\int_b^c g(x)\,dx \right)$$
You were right on all counts...

I think I got the right answer, to check, with the disks I would put $$\displaystyle /int_{-3}^9$$, or because it's y axis would I put $$\displaystyle /int^{-3}_9$$?

Since I seem to have the correct answer, I'm wagering it's the former.

Either way, how do you add [solved] to the title?