Find the volume of the solid obtained by rotating the region

Question

Find the volume of the solid obtained by rotating the region in the first quadrant bounded by the given curve about the y-axis.

y=4-(x-9)^2

Answer 1

y = 4 - x² + 18x - 81 = 18x - x² - 77 = (7 - x)(x - 11) = 0; x = 7 or x = 11

V = 2π∫₇¹¹x(18x - x² - 77)dx = 2π(6x³ - x⁴/4 - 77x²/2)₇¹¹ = 192π

Answer 2

In the first quadrant means also bounded by the x-axis.

Using shell method:

V = 2π∫^b_axf(x)dx

4-(x-9)²=0 --> -x²+18x-77 =0 --> x = 7,11

V = 2π∫¹¹₇x( -x²+18x-77)dx = 2π∫¹¹₇ -x³+18x²-77x dx = 2π[-x⁴/4+6x³-77x/2]¹¹₇=192π

Answer 3

Hi,

curve about the y-axis.

y=4-(x-9)^2

shell method:

1- solve y = 0 ====> 4-(x-9)^2 = 0 =====> x= 7 & x= 11

2- V = 2Π ∫ x f(x) dx ======> V = 2Π ∫ x[ 4-(x-9)^2 ]dx

3- V = 2Π ∫ x [ 4 - x^2 +18x -81 ]dx

4- V = 2Π ∫ [4x - x^3 +18x² - 81x ]dx now it is easy by taking integral of any term

5- V = 2Π ( 2x² -1/4 x⁴ +6x³- 81/2 x² ) now by replacing x= 7 and x= 11 and subtract you can find the volume

I hope it is useful,

Minoo

Answer 4

The given line intersects the x axis at the points x=7 and x=11. So we need to revolve that figure around y axis and find it volume.

The Volume will be

V= π∫₇¹¹[(f(x)]²dx = π∫₇¹¹ (4-(9-x)²)²dy = 2 π∫₇¹¹ (4-(9-x)²)²(9-x)cx = (128/3) π