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

Question

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?

Answer 1

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 $$ V_{\textrm{cylinder}}-V_{\textrm{curve}}=\frac{1}{2}\pi a^2 $$