finding the volume

[stop2006]

find the volume of the solid generated by revolving the region bounded by y=4?x and the lines y= 12 and x=0 about

(a) the x-axis.
(b) by y-axis.
(c) the line y=12.
(d)the line x=9.

 

[galactus]

find the volume of the solid generated by revolving the region bounded by y=4?x and the lines y= 12 and x=0 about

We can use shells with these two. With shells, the cross sections are parallel to the axis about which we are revolving.

(a) the x-axis.

$\displaystyle 2{\pi}\int_{0}^{12}y\left(\frac{y^{2}}{16}\right)dy$

(b) by y-axis.

$\displaystyle 2{\pi}\int_{0}^{9}x(12-4\sqrt{x})dx$

the line x=9

$\displaystyle 2{\pi}\int_{0}^{9}(x-9)(4\sqrt{x}-12)dx$


Try using washers on y=12

 

[BigGlenntheHeavy]

$\displaystyle a) \ x-axis, \ disc \ and \ shell.$

$\displaystyle \pi\int_{0}^{9}(144-16x)dx \ = \ 684\pi \ = \ \frac{\pi}{8}\int_{0}^{12}y^3dy.$

$\displaystyle b) \ y-axis, \ disc \ and \ shell.$

$\displaystyle \frac{\pi}{256}\int_{0}^{12}y^4dy \ = \ \frac{972\pi}{5} \ = \ 2\pi\int_{0}^{9}(12x-4x^{3/2})dx.$

$\displaystyle c) \ line \ y \ = \ 12, \ disc \ and \ shell.$

$\displaystyle \pi\int_{0}^{9}(4x^{1/2}-12)^2dx \ = \ 216\pi \ = \ \frac{\pi}{8}\int_{0}^{12}(12y^2-y^3)dy.$

$\displaystyle d) \ line \ x \ = \ 9, \ disc \ and \ shell.$

$\displaystyle \pi\int_{0}^{12}\bigg[81-\bigg(\frac{y^2}{16}-9\bigg)^2\bigg]dy \ = \ \frac{2268\pi}{5} \ = \ 2\pi\int_{0}^{9}(12-4x^{1/2})(9-x)dx.$