Question

using the shell method to find volume if you are rotating about the axis x=-5, is the radius x+5 or x-5?

Answer 1

x saiken200

Answer 2

what?

Answer 3

the radius is only x if it's rotating around the y axis itself. it's not. it's rotating around the axis x=-5. so it's -5 away from the y axis.

Answer 4

for the love of god will someone please help me!!!!!

Answer 5

i have tried this damn problem 6 different ways and cannot get the right damn answer!!!!!!!Q~!Q!@w'x

Answer 6

its x + 5

Answer 7

ok, answer's still wrong....

The region bounded by y=x8 and y=sin(πx/2) is rotated about the line x=−5.

Answer 8

\[2\pi \int\limits_{0}^{1}(x+5)(x^8-\sin(\pi x/2)\]

Answer 9

i integrate that, plug in 1-0 and get it wrong....this assignment is due in a few hours! i NEED HELP!!!

Answer 10

i am an illegal immegrant.. my nicknake is 2 bags.. i found a little bit of sheeeeeet on my ppe.. i flicked it off.. like i didnt know it was there.. im sorry...

Answer 11

how is that supposed to be an answer? i seriously need help.

Answer 12

r = k - f(x), if you are trying to assess a radius

otherwise you can just shift the setup to the y axis .... by f(x-c)

Answer 13

@saiken2009 you set up the problem as
*** \( 2\pi \int\limits_{0}^{1}(x+5)(x^8-\sin(\pi x/2)\) *****
but it is better to write it as
\[ 2\pi \int\limits_{0}^{1}(x+5)(\sin(\pi x/2)- x^8) \]
because the sin is the "top" curve

Answer 14

if you multiply it out you get
\[ 2\pi \int\limits_{0}^{1}(x+5)(\sin(\pi x/2)- x^8) \ dx \\
2\pi \int\limits_{0}^{1} x \sin\left(\frac{\pi x}{2}\right)+ 5 \sin\left(\frac{\pi x}{2}\right)-x^9 -5x^8 \ dx
\]

Answer 15

the first integral can be done using "integration by parts"
the 2nd integral is straight forward:
\[ 2\pi \int\limits 5 \sin\left(\frac{\pi x}{2}\right)=
2\pi \cdot 5 \cdot \frac{2}{\pi} \cdot -1 \cdot \cos\left(\frac{\pi x}{2}\right) =
-20 \cos\left(\frac{\pi x}{2}\right)
\]

as are the last 2:
\[ 2\pi \int -x^9= -\frac{2 \pi}{10} x^{10}\\
2\pi \int -5x^8= -\frac{10 \pi}{9} x^{9}
\]