Why is Quadrant 1 Vol x 4 Incorrect for Problem #13?

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Here on problem #13 i am getting a false solution when only taking the volume of the region in quadrant 1 and multiplying it by 4. The later photo contains the correct solution which is obtained by taking the volume of the entire region. Can someone explain to me why the former is incorrect.
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namehere said:
Here on problem #13 i am getting a false solution when only taking the volume of the region in quadrant 1 and multiplying it by 4. The later photo contains the correct solution which is obtained by taking the volume of the entire region. Can someone explain to me why the former is incorrect.
Just because you are calculating a triple integral, it doesn't mean you are calculating a volume. If the integrand was ##1##, you would be, and your method would be correct. But your integrand is ##x^2e^y##, and ##e^y## is not symmetric about ##y=0##. So you don't get the same answer for each quadrant.
 
namehere said:
Here on problem #13 i am getting a false solution when only taking the volume of the region in quadrant 1 and multiplying it by 4. The later photo contains the correct solution which is obtained by taking the volume of the entire region. Can someone explain to me why the former is incorrect.
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Your images are unreadable. Type out your work if you really want help.
 
I tried the problem myself. Is anyone else getting ##\frac{8}{3}e^{-1}##?
 
Eclair_de_XII said:
I tried the problem myself. Is anyone else getting ##\frac{8}{3}e^{-1}##?
Yes, that's correct.
 
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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