volume of solid of revolution Watch

bigmansouf
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Question:

a hemispherical bowl of internal radius 13 cm contains water to a maximum depth of 8 cm.

My attempt:
x^2+y^2=r^2
 y^2= r^2 -x^2
 y = (r^2 -x^2)^{\frac{1}{2}}
r = 13 sub it into y
 y = (13^2-x^2)^{\frac{1}{2}}= (169-x^2)^{\frac{1}{2}}

 v = \int_{0}^{8} \pi ((169-x^2)^{\frac{1}{2}})^{2}\mathrm{d  } x = \pi \left [169x - \frac{x^3}{3} \right ]_{0}^{8} = \frac{3544}{3} \pi cm^3

I was wrong but I cannot understand why the book gives answer ;  \frac{1984}{3} \pi

please help
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ghostwalker
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(Original post by bigmansouf)
Question:

a hemispherical bowl of internal radius 13 cm contains water to a maximum depth of 8 cm.

My attempt:
x^2+y^2=r^2
 y^2= r^2 -x^2
 y = (r^2 -x^2)^{\frac{1}{2}}
r = 13 sub it into y
 y = (13^2-x^2)^{\frac{1}{2}}= (169-x^2)^{\frac{1}{2}}

 v = \int_{0}^{8} \pi ((169-x^2)^{\frac{1}{2}})^{2}\mathrm{d  } x = \pi \left [169x - \frac{x^3}{3} \right ]_{0}^{8} = \frac{3544}{3} \pi cm^3

I was wrong but I cannot understand why the book gives answer ;  \frac{1984}{3} \pi

please help
You did everything correctly, except you have the wrong limits on your integral. Try drawing a diagram - where's the bottom 8cm of the bowl.
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bigmansouf
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(Original post by ghostwalker)
You did everything correctly, except you have the wrong limits on your integral. Try drawing a diagram - where's the bottom 8cm of the bowl.
thank you very much for helping me
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