# Volume of a solid of revolution: check and information

I have to calculate the volume of the solid of revolution around the $$Y$$ axis generated from the curve $$f(x) = 2x^2-4$$ between $$[0, 2]$$.

So I used the definition:

$$V_y = 2\pi\int_a^b x f(x)\ \text{d}x$$

Whence

$$v_y = 2\pi\int_0^2 x (2x^2-4)\ \text{d}x$$

But when I perform the calculation I get zero. Is there something wrong? I find it hard to believe the volume of the paraboloid obtained this way can be zero...

Also, can you please check if my formulas are correct? I wrote those ones for what concerns surface areas and volumes for solid of revolutions:

$$\text{Volume for a revolution around X axis}\quad\quad V_x = \pi\int_a^b f^2(x)\ \text{d}x$$

$$\text{Volume for a revolution around Y axis}\quad\quad V_x = 2\pi\int_a^b |x f(x)| \ \text{d}x$$

$$\text{Volume for a revolution around X axis of a region between two curves}\quad\quad V_x = \pi\int_a^b |f^2(x)- g^2(x)|\ \text{d}x$$

$$\text{Volume for a revolution around Y axis of a region between two curves}\quad\quad V_x = 2\pi\int_a^b x\cdot |f(x)- g(x)|\ \text{d}x$$

$$\text{Surface Area for a revolution around X axis} \quad \quad A_x = 2\pi \int_a^b f(x) \sqrt{1 + [f^{'}(x)]^2}\ \text{d}x$$

$$\text{Surface Area for a revolution around Y axis} \quad \quad A_x = 2\pi \int_a^b x \sqrt{1 + [f^{'}(x)]^2}\ \text{d}x$$

Also I wanted to ask you for what would be the formulas for the surface area of revolution of a region between two curves, around $$X$$ and around $$Y$$ axis.

Thank you so much, and forgive me for the many questions. I just need to check if I took good notes, and I am pretty sure I wrote down them correctly, but one never knows...

The volume of a solid of revolution generated by the region under the graph of $$f$$ over $$[a, b]$$, $$a \geq 0$$, is $$2 \pi \int_a^b |x f(x)| \,dx .$$
Notice that in our case, where $$f(x) = 2 x^2 - 4$$, $$f(x)$$ is nonnegative on $$[1, \sqrt{2}]$$ and nonpositive on $$[\sqrt{2}, 2]$$, so $$|x f(x)| = x (2x^2 - 4)$$ only for $$x \in [1, \sqrt{2}]$$, and it is the negative of that quantity for $$x \in [\sqrt{2}, 2]$$. We can evaluate the resulting integral as we would any piecewise function; in this case: $$2\pi \int_0^2 |x f(x)| \,dx = 2\pi \left[\int_0^\sqrt{2} x(2x^2 - 4) \,dx + \int_\sqrt{2}^2 - x (2 x^2 - 4) \,dx\right] .$$
Remark Since you're already observed that integral $$\int_0^2 x f(x) \,dx = 0$$, we know that both integrals on the r.h.s. of the previous display equation must have the same value.