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Find the area of the finite region enclosed by the two curves $y=x^2-5x$ and $y=5x-x^2$

Find the volume of the solid generated when the finite region between the curve $y=5x-x^2$ and the $x$-axis is rotated through three right angles about the $x$-axis leaving your answer in terms of $\pi$

I know how to do it if it was four right angles but three right angles makes it complicated.I have tried repeatedly but been unable to solve it.

Thanks.

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  • $\begingroup$ Your question does ask with four right angles $\endgroup$ – Nεo Pλατo Feb 24 '20 at 18:41
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If it's rotated by $270°$ then it forms a cross section of infinitely many quarter circle disks instead of full disks.

Hence a factor of $\dfrac{3}{4}$ comes in. Or you could think of the whole thing with a quarter cut out of it. The entire cross section is reduced by $25$ percent.

$$ V=\displaystyle \int_0^5 \dfrac{3}{4} \cdot 2\pi \cdot (x^2-5x)^2 \mathrm{dx}=156.25\pi $$

Incase you wanted the enclosed area one function is the negative of the other, so absolute area between them is twice the usual integral. The intervals are where the functions intersect basically at $x=\{0,5\}$

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