Recent content by delapcsoncruz

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...

r^2=2rsin(theta)

can you give me an example of your useful tip.. please..

the value of 'r' in 5pi/4 is -1.41 , so that is also equal to pi/4 which r is 1.41

can you please give me of an example using your said useful tip.. please..

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...

ah ok..thank you very much!

but , was my answer correct?

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..

thank you very much! :smile: ok i will revise my solution to make it more understandable... thank you again..

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...