## Volume of solid under the curve y= 7/(x^2 +5x +6)

Limits, differentiation, related rates, integration, trig integrals, etc.
chinex9a
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### Volume of solid under the curve y= 7/(x^2 +5x +6)

Volume of solid under the curve y= 7/(x^2 +5x +6) from x = 0 to x = 1 if rotated about the y-axis.

I made x the subject of the formula getting (sqrt(y+28)-5 sqrt(y))/(2 sqrt(y)) and then I found what y is when x= 0 and x=1, which were y=7/6 and y=7/12 respectively.

then I integrated using (pi*(f(y))^2*dy); pi*((sqrt(y+28)-5 sqrt(y))/(2 sqrt(y)))^2 to get the volume then I got

pi*(-7/12+105 log(2)-70 log(3)+log(128)) which is aprroximately 0.46.

I know the answer isn't correct because I checked the answers on my text book. Plz help me understand this problem

Thank you

Martingale
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### Re: Volume of solid under the curve y= 7/(x^2 +5x +6)

...

I made x the subject of the formula getting (sqrt(y+28)-5 sqrt(y))/(2 sqrt(y)) and then I found what y is when x= 0 and x=1, which were y=7/6 and y=7/12 respectively.
...
Why would you want to do that?

chinex9a
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### Re: Volume of solid under the curve y= 7/(x^2 +5x +6)

because we are rotating about the y axis

Martingale
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### Re: Volume of solid under the curve y= 7/(x^2 +5x +6)

because we are rotating about the y axis
So. Use the method of cylindrical shells.

chinex9a
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### Re: Volume of solid under the curve y= 7/(x^2 +5x +6)

Yes that is what I did right. pi*(f(y))^2 dy

Martingale
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### Re: Volume of solid under the curve y= 7/(x^2 +5x +6)

Yes that is what I did right. pi*(f(y))^2 dy
$\int\limits_{a}^{b}2\pi x f(x)dx$

chinex9a
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### Re: Volume of solid under the curve y= 7/(x^2 +5x +6)

and so are b and a 7/6 and 7/12 ?

chinex9a
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### Re: Volume of solid under the curve y= 7/(x^2 +5x +6)

Am I correct????????????

Martingale
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### Re: Volume of solid under the curve y= 7/(x^2 +5x +6)

and so are b and a 7/6 and 7/12 ?
No. a and b are 0 and 1