Question

Use the method of cylindrical shells to find the volume V generated by rotating the region bounded by the given curves about the specified axis.
y = 6x − x2, y = 8; about x = 2

Answer 1

Answer:

$\mathbf{V = [\dfrac{ 8 \pi }{3}] }$

Step-by-step explanation:

Given that:

y = 6x - x² , y = 8  about  x = 2

To find the volume of the region bounded by the curves about x = 2; we have the radius of the cylindrical shell to be x - 1, the circumference to be 2 π (x -2 ) and the height to be 6x - x² - 8

6x - x² - 8

6x - x² - 8 = 0

-x² + 6x - 8 = 0

x² - 6x + 8 = 0

(x -4) (x - 2  ) = 0

So;

x = 2 , x = 4

Thus, the region bound of the integral  are from a = 2 and b = 4

Therefore , the volume of the solid can be computed as :

$V = \int \limits ^b _a \ 2x \times f(x) \ dx$

$V = \int \limits ^4_2 2 \pi (x -2) (6x -x^2 -8) \ dx$

$V = 2 \pi \int \limits ^4_2 (6x^2 - x^3 -8x -12 x - 2x^2 +16) \ dx$

$V = 2 \pi \int \limits ^4_2 (8x^2 -x^3-20x +16) \ dx$

$V = 2 \pi \int \limits ^4_2 ( -x^3+8x^2-20x +16) \ dx$

$V = 2 \pi [\dfrac{ -x^7}{4}+\dfrac{8x^3}{3} -\dfrac{20x^2}{2} +16x]^4_2$

$V = 2 \pi [\dfrac{ -(4^4-2^4)}{4}+\dfrac{8(4^3-2^3)}{3} -\dfrac{20(4^2-2^2)}{2} +16(4-2) ]^4_2$

$V = 2 \pi [\dfrac{ -(256-16)}{4}+\dfrac{8(64-8)}{3} -10(16-4)} +16(2) ]$

$V = 2 \pi [\dfrac{ 4}{3}]$

$\mathbf{V = [\dfrac{ 8 \pi }{3}] }$

Answer 2

V = 8 pi / 3 ( pi “ 3.14 “ )