Question

help please number 41

Answer 1

You have to do it the parametric way or first re-parametrize it?

Answer 2

this is what i did

Answer 3

You are getting the general setup correctly.

Answer 4

problems finishing integration?

Answer 5

\(\color{black}{\displaystyle\int_{1}^{4} 2\pi \left(\frac{t^3}{3} +\frac{1}{2}t^{-2} \right)\sqrt{2t^{-1}+t^4+t^{-6}}dt }\)

first I would divide by \(t^3\) on the outside,
and multiply times \(t^{6}\) on the inside.

\(\color{black}{\displaystyle\int_{1}^{4} 2\pi \left(\frac{1}{3} +\frac{1}{2}t^{-5} \right)\sqrt{t^{10}+2t^{5}+1}dt }\)

Answer 6

\(\color{black}{\displaystyle\int_{1}^{4} 2\pi \left(\frac{1}{3} +\frac{1}{2}t^{-5} \right)\sqrt{\left(t^{5}+1\right)^2}dt }\)

Answer 7

yes , can you show me the steps under the radical.. i keep making mistakes lol

Answer 8

Which steps, the first when I multiplied times \(t^{6}\) ?

Answer 9

after squaring both terms you should get
\[ 4t^{-1} + t^4 +t^{-6}-2t^{-1} \]

Answer 10

I added like terms.

Answer 11

hmm how u get 4t^-1 ?
i thought it was suppose to be 4t^-2 ?

Answer 12

and after combining like terms
\[ t^4 +2 t^{-1} + t^{-6} \]
as Solomon showed, you factor out t^-6
i.e.. multiply all terms by t^6 and compensate by dividing by t^6
you get
\[ t^{-6} ( t^{10} + 2t^5 +1 ) \]
the idea is to hope for a perfect square, which is why you try to make the lowest term 1
in this case, it works out.

Answer 13

you square 2 t^(-½)

Answer 14

Marcelie, no I shouldn't. Yes, as phi explains.

Answer 15

\(\color{black}{\displaystyle \left(\frac{2}{\sqrt{t}}\right)^2=\frac{2^2}{(\sqrt{t})^2}=\frac{4}{t}=4t^{-1} }\)

Answer 16

Right?

Answer 17

oh yeah

Answer 18

yes, so we obtain

\(\color{black}{\displaystyle\int_{1}^{4} 2\pi \left(\frac{t^3}{3} +\frac{1}{2}t^{-2} \right)\sqrt{4t^{-1}+t^4-2t^{-2}+t^{-6}}dt }\)

Answer 19

then, I added like terms inside the sqrt to get

\(\color{black}{\displaystyle\int_{1}^{4} 2\pi \left(\frac{t^3}{3} +\frac{1}{2}t^{-2} \right)\sqrt{2t^{-1}+t^4+t^{-6}}dt }\)

Answer 20

then, multiplying times the so called "magic one". In other words, divided outside sqrt by \(t^{-3}\) and inside the sqrt multiplied times \(t^6\), which gives

\(\color{black}{\displaystyle\int_{1}^{4} 2\pi \left(\frac{1}{3} +\frac{1}{2}t^{-5} \right)\sqrt{2t^{5}+t^{10}+1}dt }\)

Answer 21

Interrupt me with a question, if you don't understand something ...

Answer 22

Now, inside the sqrt we have a perfect-square trinomial.

\(\color{black}{\displaystyle\int_{1}^{4} 2\pi \left(\frac{1}{3} +\frac{1}{2}t^{-5} \right)\sqrt{\left(t^5+1\right)^2}dt }\)

Answer 23

hmm i got lost where you are diving t^-3

Answer 24

because I multiply times t^6 inside the square root.

Answer 25

In other words,

\(\color{black}{\displaystyle\int_{1}^{4} 2\pi \left(\frac{t^3}{3} +\frac{1}{2}t^{-2} \right)\sqrt{2t^{-1}+t^4+t^{-6}}dt }\)

\(\color{black}{\displaystyle\int_{1}^{4} 2\pi \left(\frac{t^3}{3} +\frac{1}{2}t^{-2} \right)\times \frac{\color{blue}{t^3}}{\color{red}{t^3}}\sqrt{2t^{-1}+t^4+t^{-6}}dt }\)

\(\color{black}{\displaystyle\int_{1}^{4} 2\pi \left(\frac{t^3}{3} +\frac{1}{2}t^{-2} \right)\color{red}{\div t^3\times }\sqrt{\color{blue}{t^6}\color{blue}{\left(\color{black}{t^4+2t^{-1}+t^{-6}}\right)}}dt }\)

Answer 26

and then, this is simplified as

\(\color{black}{\displaystyle\int_{1}^{4} 2\pi \left(\frac{1}{3} +\frac{1}{2}t^{-5} \right)\sqrt{t^{10}+2t^5+1}dt }\)

Answer 27

So far so good?

Answer 28

If you have questions, you can definitely ask:)

Answer 29

lol i kinda understand but i still dont understand by the t^3

Answer 30

We want to get rid of negative exponents inside the square root, this is why we want to multiply times \(t^6\) inside the square root (which is the same as multiplying times \(t^3\) on the outside of the sqrt).

Answer 31

However, we can't just multiply outside the square-root times \(t^3\) because that completely changes the integrand. For this reason, in order to avoid "violent" changes, we multiply times \(t^3/t^3\).

Answer 32

So, the top \(t^3\) (which I colored in blue) goes inside
the sqrt as \(t^6\) (because \(t^3=\sqrt{t^6}\)), and the bottom \(t^3\)
we distribute into the parenthesis of \((\)t^3/3 + 1/2t\(^{-2}\)\()\).

Answer 33

For this reason we did:

\(\color{black}{\displaystyle\int_{1}^{4} 2\pi \left(\frac{t^3}{3} +\frac{1}{2}t^{-2} \right)\times \frac{\color{blue}{t^3}}{\color{red}{t^3}}\sqrt{2t^{-1}+t^4+t^{-6}}dt }\)

Answer 34

which as I wrote results in \(\color{black}{\displaystyle\int_{1}^{4} 2\pi \left(\frac{1}{3} +\frac{1}{2}t^{-5} \right)\sqrt{t^{10}+2t^5+1}dt }\)

Answer 35

Better?

Answer 36

yeah i get it now x)

Answer 37

Alright!

Answer 38

Now, with a little closer attention to the part inside sqrt, we see that it's a perfect square trinomial.

Answer 39

\(\color{black}{\displaystyle\int_{1}^{4} 2\pi \left(\frac{1}{3} +\frac{1}{2}t^{-5} \right)\sqrt{t^{10}+2t^5+1}dt }\)

\(\color{black}{\displaystyle\int_{1}^{4} 2\pi \left(\frac{1}{3} +\frac{1}{2}t^{-5} \right)\sqrt{(t^5+1)^2 }dt }\)

\(\color{black}{\displaystyle\int_{1}^{4} 2\pi \left(\frac{1}{3} +\frac{1}{2}t^{-5} \right)\left(t^5+1\right)dt }\)

Answer 40

Remark:

In general, \(\color{black}{\displaystyle \sqrt{\left[f(x)\right]^2}=\left|f(x)\right| }\), however, in our case, knowing that \(\color{black}{\displaystyle t\ge0 }\),
we have \(\color{black}{\displaystyle t^5+1\ge 1 }\), and since this quantity is always positive anyway, the
absolute value bars are omitted.

Answer 41

Next, is to expand and finish the integration with nothing more complex than the power rule.

Question? Ask!
This is your task! :)

Answer 42

okay got it ty :D

Answer 43

No problem:)