You are using an out of date browser. It may not display this or other websites correctly.
You should upgrade or use an alternative browser.
You should upgrade or use an alternative browser.
finding the volume
- Thread starter stop2006
- Start date
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
Senior Member
- Joined
- Mar 8, 2009
- Messages
- 1,577
\(\displaystyle a) \ x-axis, \ disc \ and \ shell.\)
\(\displaystyle \pi\int_{0}^{9}(144-16x)dx \ = \ 684\pi \ = \ \frac{\pi}{8}\int_{0}^{12}y^3dy.\)
\(\displaystyle b) \ y-axis, \ disc \ and \ shell.\)
\(\displaystyle \frac{\pi}{256}\int_{0}^{12}y^4dy \ = \ \frac{972\pi}{5} \ = \ 2\pi\int_{0}^{9}(12x-4x^{3/2})dx.\)
\(\displaystyle c) \ line \ y \ = \ 12, \ disc \ and \ shell.\)
\(\displaystyle \pi\int_{0}^{9}(4x^{1/2}-12)^2dx \ = \ 216\pi \ = \ \frac{\pi}{8}\int_{0}^{12}(12y^2-y^3)dy.\)
\(\displaystyle d) \ line \ x \ = \ 9, \ disc \ and \ shell.\)
\(\displaystyle \pi\int_{0}^{12}\bigg[81-\bigg(\frac{y^2}{16}-9\bigg)^2\bigg]dy \ = \ \frac{2268\pi}{5} \ = \ 2\pi\int_{0}^{9}(12-4x^{1/2})(9-x)dx.\)
\(\displaystyle \pi\int_{0}^{9}(144-16x)dx \ = \ 684\pi \ = \ \frac{\pi}{8}\int_{0}^{12}y^3dy.\)
\(\displaystyle b) \ y-axis, \ disc \ and \ shell.\)
\(\displaystyle \frac{\pi}{256}\int_{0}^{12}y^4dy \ = \ \frac{972\pi}{5} \ = \ 2\pi\int_{0}^{9}(12x-4x^{3/2})dx.\)
\(\displaystyle c) \ line \ y \ = \ 12, \ disc \ and \ shell.\)
\(\displaystyle \pi\int_{0}^{9}(4x^{1/2}-12)^2dx \ = \ 216\pi \ = \ \frac{\pi}{8}\int_{0}^{12}(12y^2-y^3)dy.\)
\(\displaystyle d) \ line \ x \ = \ 9, \ disc \ and \ shell.\)
\(\displaystyle \pi\int_{0}^{12}\bigg[81-\bigg(\frac{y^2}{16}-9\bigg)^2\bigg]dy \ = \ \frac{2268\pi}{5} \ = \ 2\pi\int_{0}^{9}(12-4x^{1/2})(9-x)dx.\)