Why Is My Calculus Volume Integration Result Incorrect?

kira137
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I've been trying this equation for few hours now and I can't seem to get the right answer..
answer is suppose to be 640(pi)/3

Homework Statement

Find the volume of the solid of revolution obtained by rotating the region bounded by the curves y=x^(2)+1 and y = 9-x^2 around the x-axisThe attempt at a solution
``2
Pi S (9-x^2)^2 - (x^2+1)^2 dx
``-2

``2
Pi S (81-2x^2+x^4-x^4-2x^2-1) dx
``-2

Pi[80x - (4x^3)/3] (x=-2 to 2)

= (448(pi)/3) - (-448(pi)/3) = 896(pi)/3

that's what I keep getting..
and if that isn't the correct answer for rotating around x=0, isn't it suppose to be the answer for when rotating around y=1?

thank you in advance
 
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kira137 said:
I've been trying this equation for few hours now and I can't seem to get the right answer..
answer is suppose to be 640(pi)/3

Homework Statement

Find the volume of the solid of revolution obtained by rotating the region bounded by the curves y=x^(2)+1 and y = 9-x^2 around the x-axis


The attempt at a solution
``2
Pi S (9-x^2)^2 - (x^2+1)^2 dx
``-2

``2
Pi S (81-2x^2+x^4-x^4-2x^2-1) dx
``-2

Pi[80x - (4x^3)/3] (x=-2 to 2)

= (448(pi)/3) - (-448(pi)/3) = 896(pi)/3

that's what I keep getting..
and if that isn't the correct answer for rotating around x=0, isn't it suppose to be the answer for when rotating around y=1?

thank you in advance

Check again your algebra when you expand the squares inside the integration.

Cheers -- sylas
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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