Question

Need help with evaluating this integral:

(16-r^2)^1/2r dr dtheta.

Answer 1

\[\int\limits_{0}^{2\pi}\int\limits_{0}^{2}\sqrt{16-r^2}r dr d \theta\]

Answer 2

This problem is worked out in my book, theres a certain part where I'm stuck on how they evaluated the problem i'll post that up

Answer 3

\[-1/3\int\limits_{0}^{2\pi}(16-r^2)^{3/2} \theta \]
Where are they getting the squareroot of 3 here?
\[-1/3 \int\limits_{0}^{2\pi}(24\sqrt{3}-64)d \theta\]

Answer 4

I understand the 24 and the 64...

oh and I forgot to put the inner limits in brackets at the end of the first equation i typed in.

Answer 5

firm producing hokey sticks has a production function given by Q=2√(K.L) Where: K= Capital inputs L= Labour inputs In the short run, the firm amounts of capital equipments is Fixed at K=100. The rental rate for capital is V=1.00 Shillings and the wage rate for L=4.00 shillings. Calculate the firms short run costs functions as well as the short run average costs Functions. What is the firms short run marginal costs. any one who can help

Answer 6

\[\int\limits_{0}^{2\pi}\int\limits_{0}^{2}\sqrt{16-r^2}r dr d \theta\] first
\[\int\limits_{0}^{2}\sqrt{16-r^2}r dr\] a u sub gives
\[-\frac{1}{3}(16-r^2)^{\frac{3}{2}}\]

Answer 7

yep, I follow you so far

Answer 8

replace x by 2 and get
\[-\frac{1}{3}(16-2^2)^{\frac{3}{2}}\]
\[-\frac{1}{3}\sqrt{12}^3\]
\[-\frac{1}{3}\times 24\sqrt{3}\]
\[-8\sqrt{3}\]

Answer 9

replace x by 0 and get
\[-\frac{1}{3}(16)^{\frac{3}{2}}\] etc

Answer 10

ok

Answer 11

thanks

Answer 12

yw

Answer 13

wait, now I'm confused about the 24. Where is that coming from?

Answer 14

Do you still don't know where the 24√3 comes from?

Answer 15

Yeah

Answer 16

\[\sqrt{12}^{3} = (2\sqrt{3})^{3} = 8*3*\sqrt{3} = 24\sqrt{3}\]

Answer 17

That other three is coming from -1/3 right?

Answer 18

if so then i think i get it

Answer 19

No i'm wrong, cause then it would have been -24squareroot of 3

Answer 20

? not following you

Answer 21

I'm a bit confused on your answer, I know you got 8 from 2^3 but how are you getting the 8*3 part?

Answer 22

you should understand how they get to this point now
\[-1/3\int\limits_{0}^{2\pi}(24\sqrt{3}-64) d \theta\]

Answer 23

\[\sqrt{3}^{3} = \sqrt{3}*\sqrt{3}*\sqrt{3} = \sqrt{9}*\sqrt{3}=3*\sqrt{3}\]

Answer 24

oh ok now I see. Thanks.

Answer 25

no problem