ok i'll do that. you are right, 't' takes the place of either dx or dy, it is the thickness. thank you very much for your reply and time. appreciate it. thanks.. :))
Homework Statement
Find the volume of the first quadrant region bounded by x=y-y3, x=1 and y=1 that is revolved about the line x=1.
The Attempt at a Solution
dV=∏R2t
where :
t=dy
R=1-(y-y3)
=1-y+y3
so..
dV=∏(1-y+y3)2dy
dV=∏(1-2y+y2+2y3-2y4+y6)dy
V=∏∫ from 0 to 1 of...
what is the area inside the graph of r=2sinθ and outside the graph of r=sinθ+cosθ?
so i compute for the values of 'r',... but, i only got one intersection point which is (45°, 1.41).
there must be two intersection points right? but I've only got one. what shall i do?
i cannot compute for...
Homework Statement
Find the area between y= 1/(x2+1) and the x-axis, from x=0 to x=1
The Attempt at a Solution
so when x=0, y=1
and when x=1, y=1/2
next i plot the points, so the intersection of the given equation is (0,1) and (1,1/2)
Yh= Y-higher= 1/(x2+1)
Yl= Y-lower= 0
the...
I need some help on the interpretation of our laboratory experiment in physics.
In our lab experiment we charged and uncharged a capacitor. And here are the things we observe.
1. when one bulb was connected in a circuit across a 4.5 V potential source, the bulb turned on with a bright...
ok this what I've got...
=xax - ∫axdx/lna
=xax - 1/lna∫axdx
=xax - (1/lna) (ax/lna)
=xax - ax/2lna
is that right that (lna)(lna) = 2lna?
or it is just (lna)(lna) = (lna)(lna)
∫xax
u=x
du=dx
dv=axdx
v=ax/lna
= xax - ∫axdx/lna
is my solution right?
my problem now is how to integrate the expression xax - ∫axdx/lna
please help..
i solve it and this is my solution..
∫ (x+2)dx / sqrt (4x-x2)
let u= 4x -x2
du=-2(x-2)dx
so now the numerator will become [(x-2)+4]dx
= ∫ [(x-2)+4]dx / sqrt (4x-x2)
= ∫ (x-2)dx / sqrt (4x-x2) + ∫ 4dx / sqrt (4x-x2)
let u = 4x -x2
du = -2(x-2)dx...