How do you find the surface area of the solid obtained by rotating about the $x$ -axis the region bounded by $9x=y^2+18$ on the interval $2<=x<=6$ ?

Question

10828 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2014-09-20T13:34:26

The surface area can be found by

$S=2piint_0^6 ysqrt{1+{4y^2}/81}dy=49pi$

Let us look at some details.

$9x=y^2+18 Leftrightarrow x=y^2/9+2 Leftrightarrow y=pm3sqrt{x-2}$

By taking the derivative,

$dx/dy={2x}/9$

As x goes from 2 to 6, y goes from 0 to 6.

Since the surface area of revolution can be found by

$S=2piint_a^b y\sqrt{1+(dx/dy)^2}dy$ ,

we have

$S=2piint_0^6 ysqrt{1+{4y^2}/81}dy$

by substitution $u=1+{4y^2}/81$ . $Rightarrow {du}/dy={8y}/81 Rightarrow dy=81/{8y}du$ ,
As y goes from 0 to 6, u goes from 1 to 25/9.

$=2pi int_1^{25/9}ysqrt{u}{81}/{8y}du$

by simplifying,

$={81pi}/4 int_1^{25/9}u^{1/2}du$

$={81pi}/4[2/3u^{3/2}]_1^{25/9}$

$={27pi}/2[(25/9)^{3/2}-1]=49pi$

How do you find the surface area of the solid obtained by rotating about the xx-axis the region bounded by 9x=y^2+189x=y^2+18 on the interval 2<=x<=62<=x<=6 ?