Please helpp :) (1 Viewer)

[umm what]

plz do (26)
thnxx

 

[Cl324]

56/3pi i think

 

[Timske]

y = x + 2 , the volume is bounded by x = 0 and x = 2

v = pi*int x + 2
= pi [ x^2/2 + 2x]

sub in 2 and 0

= pi [(2^2/2 + 2(2)) ] - [(0^2/2 + 2(0) ]
= pi [2 + 2] - 0
= 4 pi cubic units

oops meant to be 2 + 4 = 6
so its 6pi

 

[deswa1]

<a href="http://www.codecogs.com/eqnedit.php?latex=V=\pi\int_{0}^{2}y^2dx\\ V=\pi\int_{0}^{2}(x@plus;2)^2dx \\V=\pi\int_{0}^{2}x^2@plus;4x@plus;4dx\\ V=\left [ \frac{x^3}{3}@plus;2x^2@plus;4x \right ]^{2}_{0}\\ V=\frac{56\pi}{3}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?V=\pi\int_{0}^{2}y^2dx\\ V=\pi\int_{0}^{2}(x+2)^2dx \\V=\pi\int_{0}^{2}x^2+4x+4dx\\ V=\left [ \frac{x^3}{3}+2x^2+4x \right ]^{2}_{0}\\ V=\frac{56\pi}{3}" title="V=\pi\int_{0}^{2}y^2dx\\ V=\pi\int_{0}^{2}(x+2)^2dx \\V=\pi\int_{0}^{2}x^2+4x+4dx\\ V=\left [ \frac{x^3}{3}+2x^2+4x \right ]^{2}_{0}\\ V=\frac{56\pi}{3}" /></a>

 

[deswa1]

y = x + 2 , the volume is bounded by x = 0 and x = 2

v = pi*int x + 2
= pi [ x^2/2 + 2x]

sub in 2 and 0

= pi [(2^2/2 + 2(2)) ] - [(0^2/2 + 2(0) ]
= pi [2 + 2] - 0
= 4 pi cubic units

oops meant to be 2 + 4 = 6
so its 6pi

You made a mistake in your first line. It should be the integral of (x+2)^2 not (x=2)

 

[umm what]

thanks brahh

<a href="http://www.codecogs.com/eqnedit.php?latex=V=\pi\int_{0}^{2}y^2dx\\ V=\pi\int_{0}^{2}(x@plus;2)^2dx \\V=\pi\int_{0}^{2}x^2@plus;4x@plus;4dx\\ V=\left [ \frac{x^3}{3}@plus;2x^2@plus;4x \right ]^{2}_{0}\\ V=\frac{56\pi}{3}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?V=\pi\int_{0}^{2}y^2dx\\ V=\pi\int_{0}^{2}(x+2)^2dx \\V=\pi\int_{0}^{2}x^2+4x+4dx\\ V=\left [ \frac{x^3}{3}+2x^2+4x \right ]^{2}_{0}\\ V=\frac{56\pi}{3}" title="V=\pi\int_{0}^{2}y^2dx\\ V=\pi\int_{0}^{2}(x+2)^2dx \\V=\pi\int_{0}^{2}x^2+4x+4dx\\ V=\left [ \frac{x^3}{3}+2x^2+4x \right ]^{2}_{0}\\ V=\frac{56\pi}{3}" /></a>