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Compute the area of that part of plane $x+y+z=2a$ which lies in the first octant and is bounded by the cylinder $x^2+ y^2 =a^2$

Now $z = 2a -x -y$ ,

$\dfrac{\partial{z}}{\partial{x}} = -1$

$\dfrac{\partial{z}}{\partial{y}} = -1$ , $ds = \sqrt{3}dxdy$

Area is given by : $\int\int_{A} \sqrt{3}dxdy$ where

A is the area of $x^2+ y^2 =a^2$ in first quadrant.

so Surface area = $\dfrac{\sqrt{3}\pi a^2}{4}$. However the answer in my book is $\dfrac{3\pi a^2}{4}$ .

can anyone tell me what is wrong with my solution ? and why is my answer is incorrect ?

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2 Answers 2

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Another way:

A = πa²/4, α = angle of planes z = 0 and x + y + z = 2a

cos α = √3/3

then SA = A/cos α = πa²√3/4

I would say, Your result is correct.

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  • $\begingroup$ nice answer! thanks for the geometrical intution. $\endgroup$
    – zeroflank
    Aug 19, 2019 at 11:39
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Your answer is fine.

One way to verify is to look at the area of the whole ellipse inside the cylinder. It is not hard to find that the long axis of the ellipse is $\sqrt{3}a$. So, it’s area is $\sqrt{3}\pi a^2$. The answer should be a quarter of that, i.e. $\sqrt{3}\pi a^2/4$.

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