A Question about Volume of the object

Question

A Volume is described as follows:

1. the base is the region bounded by $x=-y^2+10y+25$ and $x=y^2-22y+135;$

2. Every cross section perpendicular to the $y-$ axis is a semi-circle

Find the volume of the object.

My attempt:

now $-y^2+10y+25=y^2-22y+135$

then $2y^2-32y+110=0 \Rightarrow y^2-16y+55=0\Rightarrow y= 5, 11$

Answer 1

Yes, the two parabolas intersect where y= 5 and y= 11. Further, when y= 5, x= -25+ 50+ 25= 50 and when y= 11, x= -121+ 110+ 25= 14. That is, the two parabolas intersect at (50, 5) and (14, 11).

Also, $-y^2+ 10y+ 25= -(y^2- 10y+ 25- 25)+ 25= -(y- 5)^2+ 50$. The first parabola has vertex at (50, 5). $y^2- 22y+ 135= (y^2- 22y+ 121- 121)+ 135= (y- 11)^2+ 14$. The second parabola has vertex at (14, 11). The intersections of the two parabolas are at their vertices! That means we don't need to break the region of integration into parts.

If we take the integral with respect to y, it has to go from 5 to 11 and, for each y, the integrand is a semi-circle with diameter equal to the x-distance between the two parabolas, $(-y^2+ 10y+ 25)- (y^2- 22y+ 135)= -2y^2+ 32y- 114$. The radius of the semi-circle is half that. The integrand is $\pi\frac{(2y^2- 32y+ 114)^2}{4}$. The integral is $\frac{\pi}{4}\int_5^{11} (2y^2- 32y+ 114)^2 dy= \pi\int_5^{11} (y^2- 16y+ 57)^2 dy$

Answer 2

Hint:

$\frac{1}{2}\int\limits_a^b \pi r^2\,dy$ where a=5;b=11

$r=\frac{(x_1-x_2)}{2}$: $x_1$ belongs to 1st equation and $x_2$ belongs to 2nd equation