hitung volume benda putar yang dibatasi oleh kurva kurva berikut: y=9-x^2 dan y=x+3 diputar mengelilingi sumbu x 360°

[tex]\displaystyle y_1=y_2\\x+3=9-x^2\\x^2+x-6=0\\(x+3)(x-2)=0\\x=2\vee x=-3\\\\V=\pi\int\limits^{x_2}_{x_1}y_2^2-y_1^2\,dx\\V=\pi\int\limits^{2}_{-3}(9-x^2)^2-(x+3)^2\,dx\\V=\pi\int\limits^{2}_{-3}81-18x^2+x^4-x^2-6x-9\,dx\\V=\pi\int\limits^{2}_{-3}x^4-19x^2-6x+72\,dx\\V=\pi\left(\left\frac15x^5-\frac{19}{3}x^3-3x^2+72x\right|^{2}_{-3}\right)\\V=\pi\left(\frac{2^5-(-3)^5}5-\frac{19(2^3-(-3)^3)}{3}-3(2^2-(-3)^2)+72(2-(-3))\right)\\V=\pi\left(\frac{32+243}5-\frac{19}{3}(8+27)-3(4-9)+72(2+3)\right)[/tex]
[tex]\displaystyle V=\pi\left(\frac{275}5-\frac{19}{3}(35)-3(-5)+72(5)\right)\\V=\pi\left(55-221\frac{2}{3}+15+360\right)\\V=\pi\left(430-221\frac{2}{3}\right)\\V=\pi\left(208\frac{1}{3}\right)\\\boxed{\boxed{V=208,33\pi\,\text{sV}}}[/tex]

Titik potong (sebagai batas integral)
y = y
x + 3 = 9 - x^2
x^2 + x - 6 = 0
(x - 2)(x + 3) = 0
x = 2 atau x = -3

V = π int [(9 - x^2)^2 - (x + 3)^2] dx
= π int [81 - 18x^2 + x^4 - (x + 3)^2 ] dx
= π [81x - 6x^3 + 1/5 x^5 - 1/3 (x + 3)^3] => batas -3 sampai 2
= π [(81(2) - 6(2)^3 + 1/5(2)^5 - 1/3 (2 + 3)^3) - (81(-3) - 6(-3)^3 + 1/5 (-3)^5 - 1/3 (-3 + 3)^3)]
= π [162 - 48 + 32/5 - 125/3 - (-243 + 162 - 243/5 - 0)]
= π [114 + 32/5 - 125/3 + 243 - 162 + 243/5]
= π [195 + 275/5 - 125/3]
= π [195 + 55 - 125/3]
= π [195 + (165 - 125)/3]
= π [195 + 40/3]
= π [195 + 13 1/3]
= 208 1/3 π

= 625/3 π