Please help with this question (1 Viewer)

kpad5991 · Jan 20, 2018

What is the volume of the solid generated when the area of the curve y=3-3^x is rotated about the x axis

HeroWise · Jan 20, 2018

Any ordinates given or assuming first quadrant?

Use the powerful Blue Eyes White logs pls don't kill me

fan96 · Jan 20, 2018

kpad5991 said:
What is the volume of the solid generated when the area of the curve y=3-3^x is rotated about the x axis

Infinity.

If bounds are given, use the formula

$V = \pi \int^a_b y^2\, dx$

and use

$k = e^{\log k}$ (for all $k > 0$ )

(remember to give your answers in cubic units!)

HeroWise · Jan 20, 2018

Well to summ wat the above posts states is the

x = e ^ (lnx)

is the same thing right?

didnt finish it fully, should be right if i didnt make silly errors, but you want to reach there

dan964 · Jan 21, 2018

Herowise one of your coefficients is not correct.

HeroWise said:
Well to summ wat the above posts states is the

x = e ^ (lnx)

is the same thing right?

didnt finish it fully, should be right if i didnt make silly errors, but you want to reach there

that is some funky integration going on there:

Might be easier to use change of base before integrating. Also maybe leave the exponent as x rather than x+1 before integrating. Less likely to make mistakes that way.
I'll do an improper integral here:
$\displaystyle\int{9-6 e^{x \ln 3} + e^{x \cdot 2\ln 3}$

$= 9x - \frac{6}{\ln 3} e^{x \ln 3} + \frac{1}{2 \ln 3} e^{2x \ln 3} + C$

$= 9x - \frac{2}{\ln 3} 3^{x+1} + \frac{1}{2 \ln 3} 3^{2x}+C$

or

$\displaystyle\int{9-2 e^{(x+1) \ln 3} + e^{x \cdot 2\ln 3}$

$=\displaystyle\int{9-2 e^{\ln 3} \cdot e^{x \ln 3} + e^{x \cdot 2\ln 3} + C$

$= 9x - \frac{2}{\ln 3} e^{\ln 3} e^{x \ln 3} + \frac{1}{2 \ln 3} e^{2x \ln 3} + C$

$$since $ e^{\ln 3} = 3$

$= 9x - \frac{6}{\ln 3} e^{x \ln 3} + \frac{1}{2 \ln 3} e^{2x \ln 3}+C$

$= 9x - \frac{2}{\ln 3} 3^{x+1} + \frac{1}{2 \ln 3} 3^{2x}+C$

This is the improper integral. The integral with limits is done in a similar fashion.

HeroWise · Jan 21, 2018

Pfft i feel dumb, i see where i went wrong, thanks a lot anyway @dan964

EDIT: Wait, no i got the same result as yours @dan964, but i didnt simplify it. if i had, it would come down to your answer, and the ordinates, as i said, im geussing the question is asking for first quadrant, i remember a similar question while going over the Cambridge book, so i just "geussed". That isnt really good but i think the working should suffice the OP

kpad5991 · Jan 21, 2018

dan964 said:
Herowise one of your coefficients is not correct.

that is some funky integration going on there:

Might be easier to use change of base before integrating. Also maybe leave the exponent as x rather than x+1 before integrating. Less likely to make mistakes that way.
I'll do an improper integral here:
$\displaystyle\int{9-6 e^{x \ln 3} + e^{x \cdot 2\ln 3}$

$= 9x - \frac{6}{\ln 3} e^{x \ln 3} + \frac{1}{2 \ln 3} e^{2x \ln 3} + C$

$= 9x - \frac{2}{\ln 3} 3^{x+1} + \frac{1}{2 \ln 3} 3^{2x}+C$

or

$\displaystyle\int{9-2 e^{(x+1) \ln 3} + e^{x \cdot 2\ln 3}$

$=\displaystyle\int{9-2 e^{\ln 3} \cdot e^{x \ln 3} + e^{x \cdot 2\ln 3} + C$

$= 9x - \frac{2}{\ln 3} e^{\ln 3} e^{x \ln 3} + \frac{1}{2 \ln 3} e^{2x \ln 3} + C$

$$since $ e^{\ln 3} = 3$

$= 9x - \frac{6}{\ln 3} e^{x \ln 3} + \frac{1}{2 \ln 3} e^{2x \ln 3}+C$

$= 9x - \frac{2}{\ln 3} 3^{x+1} + \frac{1}{2 \ln 3} 3^{2x}+C$

This is the improper integral. The integral with limits is done in a similar fashion.

This is what I got you will see the question from the textbook and my solutions I can't seem to get 13c. The books solution is pi(9-8/log3) u^3
https://imgur.com/gallery/u5PCA

Can someone please write the full solution to 13c

HeroWise · Jan 22, 2018

Can you tell us the ordinates then, because as the working out provided by both @dan964 and i are elligible, so just put in the ordinates like normal integration, or provide us with ordinates; thatd be helpful for us ty

or send a pic of the whole excercise

fan96 · Jan 22, 2018

$V = \pi \int^1_0 (3-3^x)^2 \, dx$

$= \pi \int^1_0 9 - 6(3^x) + 3^{2x}\, dx$

$= \pi \int^1_0 9 - 6e^{x \log 3} + e^{2x \log 3}\, dx$

$= \pi \left[ 9x - \frac{6}{\log 3}e^{x \log 3} + \frac{1}{2 \log 3}e^{2x \log 3} \right]^1_0$

$= \pi \left[\left(9 - \frac{6}{\log 3}e^{\log 3}+ \frac{1}{2 \log 3} e^{2 \log 3}\right)-\left(0-\frac{6}{\log 3} + \frac{1}{2 \log 3}\right) \right]$

Remember, $e^{\log x} = x,\, x > 0$

$= \pi \left(9 - \frac{18}{\log 3}+ \frac{9}{2 \log 3}+\frac{6}{\log 3} - \frac{1}{2 \log 3}\right)$

$= \pi \left(9 - \frac{12}{\log 3}+ \frac{8}{2 \log 3}\right)$

$= \pi \left(9 - \frac{12}{\log 3}+ \frac{4}{\log 3}\right)$

$= \pi \left(9 - \frac{8}{\log 3}\right) \text{u}^3$

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kpad5991

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