a challenging Volume question(Integration)

Steps are as follows:

圖片參考：http://i238.photobucket.com/albums/ff245/chocolate328154/Maths303.jpg?t=1215941791

15a.
Let the equation of circle be x^2+y^2=R^2.
x^2=R^2-y^2
V
=∫(from R-h to R)(pi)x^2dy
=∫(from R-h to R)(pi)(R^2-y^2)dy
=(pi)[(R^2)y-(y^3)/3](from R-h to R)
=(pi){R^3-(R^3)/3-[R^3-(R^2)h]+[R^3-3(R^2)h+3R(h^2)-h^3]/3}
=(pi)[R^3-(R^3)/3-r^3+(R^3)/3+(R^2)h-(R^2)h+(h^2)(3R-h)/3]
=(pi)(h^2)(3R-h)/3

15b.
By the result of (a), substitute x=3, R=5, we have y=4.

Volume of the whole sphere=4(pi)(5^3)/3=500(pi)/3
Volume of the hole
=(pi)(4^2)(3*5-4)/3+(pi)(4^2)(2)(5-4)
=28(pi)/3+72(pi)
=244(pi)/3
∴Remaining volume
=500(pi)/3-244(pi)/3
=256(pi)/3

呵呵。

先計出相應的x=f(y)

令圓心在原點

x^2+y^2=R^2
x=√(R^2-y^2)

體積
=π∫(R^2-y^2) dy [from R to R-h]
= π(R^2y-y^3/3) [from R to R-h]
=π(R^3-R^3/3-R^2(R-h)+(R-h)^3/3)
=(π/3)(3R^2h-3R^2h+3Rh^2-h^3)
=(π/3)(3Rh^2-h^3)
==(πh^2/3)(3R-h)