Find the volume of $\displaystyle y=4-x^2$ bounded by y=0 and y=4
Is this correct?
$\displaystyle y\!:\;\;V \;=\;\pi\int^4_04-y\,.dy$
$\displaystyle V \;=\;\pi\ (4y-\frac{y^2}{2}) \bigg]^4_0$
$\displaystyle V \;=\;8\pi\ units^3$
Find the volume of $\displaystyle y=4-x^2$ bounded by y=0 and y=4
Is this correct?
$\displaystyle y\!:\;\;V \;=\;\pi\int^4_04-y\,.dy$
$\displaystyle V \;=\;\pi\ (4y-\frac{y^2}{2}) \bigg]^4_0$
$\displaystyle V \;=\;8\pi\ units^3$
yes your answer is correct. you could also use the method of cylindrical shells to find the volume
x will go from x = 0 to x = 2
the integral would be
$\displaystyle V \;=2\;\pi\int^2_0x(4-x^2)\,dx$
$\displaystyle V \;=2\;\pi\int^2_04x-x^3\,dx$
$\displaystyle V \;=2\;\pi\ (2x^2-\frac{x^4}{4}) \bigg]^2_0$
$\displaystyle V \;=2\;\pi\ (8-4)$
$\displaystyle V\;=8\;\pi$