# Volume of $y=x^2$ bounded by $y=a$ with substraction

I'm trying find the volume of the volume of $$y=x^2$$ rotated around the $$y$$ axis and bounded by $$y=a$$. I'm following MIT 18.01 and professor solves the problem by slicing the shape to shells. To be more exact $$y=x^2, y=a\\One \ slice = dV = (2\pi x)(a-x^2)dx\\V=\frac{\pi }{2} a^{2}$$ But I think there is an easier solution. Why don't we just calculate the volume of the cylinder with height $$a$$ and radius = $$\sqrt{a}$$ (as $$y=x^2$$) and subtract the area under the $$y=x^2$$ curve rotated around the y axis?$$y=x^2, y=a\\V_{cylinder} = \pi\cdot\sqrt{a}^2\cdot a=\pi\cdot a^2\\Area\ under\ x^2 =A_{curve}=\int ^{\sqrt{a}}_{0} x^{2} dx\\V_{curve}=A_{curve}\cdot2\pi\\V=V_{cylinder}-V_{curve}=\pi(a^2-\frac{2}{3}a^{3/2})$$Why they do not equal the same thing?

• Althought the idea is nice, it is not well executed, because $V_{curve} = A_{curve}\cdot 2\pi$ is not quite correct. Commented Feb 20, 2021 at 18:51
• @FormerMath Could you elaborate bit more on why it is not correct. Intuitively I thought that if I multiply area with the perimeter I would get the rotated volume. Commented Feb 20, 2021 at 18:59
• A very simple argument could be the following: the rectangle of lenght 2 and height 1 gives different volume depending on the rotation axis. But they have the same area, so the formula can't be true. Commented Feb 20, 2021 at 19:22
• That's reasonable, but still I can't explain it... Commented Feb 20, 2021 at 19:47

The way you calculated $$V_{\textrm{curve}}$$ is incorrect. There is no such formula $$V_{\textrm{curve}}=A_{\textrm{curve}}\cdot 2\pi$$,

Instead, if you use the shell method: $$V_{\textrm{curve}}=\int_{0}^{\sqrt{a}}2\pi x\cdot x^2\ dx=\frac{1}{2}\pi a^2$$ so that $$V_{\textrm{cylinder}}-V_{\textrm{curve}}=\frac{1}{2}\pi a^2$$

• I got it. Then is my idea right? Commented Feb 20, 2021 at 19:03
• @log101: Your idea is not right: "The way you calculated 𝑉curve is incorrect."
– user9464
Commented Feb 20, 2021 at 19:05
• I mean if you substract $V_{curve}$ from the cylinder you get the same result? Commented Feb 20, 2021 at 19:36
• @log101: yes, that is correct.
– user9464
Commented Feb 20, 2021 at 19:37
• @log101: $A_{curve}\cdot 2\pi$ is not an analog of "area of base times height".
– user9464
Commented Feb 20, 2021 at 19:51