Calculate the volume of the solid

Question

Which is generated from the common part of $$ y =x^2$$ and $$y^2=8x$$ as it rotates around $0y$. Limits of the integral should be from $0$ to $4$ right? Shouldn't the integral be $$π\int_0^4(x^2 - \sqrt{8x})^2dx$$

Answer 1

Scale misleading

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Answer

At $y$ distance from origin we take a thin vertical ring of thickness $dy$. Now volume of this ring is: $$dV=\pi(x_o^2-x_i^2)dy$$ where $r_i$ and $r_o$ are inner and outer radii. Now inner radii will be on blue curve such that: $$8x_i=y^2$$ And the outer radii: $$x_o^2=y\implies x_o=+\sqrt y$$ Note the "+" because we're for the time being taking the I^st quadrant in consideration. Now $$dV=\pi(y-y^4/64)dy$$ They intersect at $k=h^2,k^2=8h\implies h=2,k=4$ And: $$V=\int_0^4\pi(y-y^4/64)dy=\frac{24\pi}5$$