Volume by disk perpendicular to $x$ axis

Question

Find the volume of the solid formed when region enclosed by $y=x^{1/2},\quad y=6-x$ and $y=0$ revolves around $x$ axis.

Can I use the equation for finding solids by splitting the solid into 2 parts?

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Answer 1

$y=x^{1/2},\quad y=6-x$ , so put $x^{1/2} =6-x$

square both sides and get to $x^2 -13x +36 =0$

solutions are 4 and 9, use $x=4$ and $x=0$ for the limits in the volume formula $V=\int \pi y^2 dx$ for curve $y=x^{1/2}$.

Then similar for the line equation, between 4 and 6. Do separately then add.

Draw a diagram to see the $x=6$ point and why $x=4$ was used, not $x=9$.

Answer 2

To calculate area from known shapes

$$ 4\cdot2\cdot\dfrac23+2\cdot 2\cdot\dfrac12 $$