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Find the area of the region bounded by paraboloid $z = x^2 + y^2 $ lies below $z = 4$ and in the first octant.

Where I am going wrong? What is the correct area?

My work:

$A = \int \int_D \sqrt{ \left (\dfrac{\partial z}{\partial x} \right ) + \left (\dfrac{\partial z}{\partial y} \right ) + 1} \ dA $

$z = x^2+y^2$

$z = 4$

$x^2+y^2 =4$

$\left (\dfrac{\partial z}{\partial x} \right )^2 + \left (\dfrac{\partial z}{\partial y} \right ) ^2+ 1 = 4(x^2+y^2) +1$

$\left (\dfrac{\partial z}{\partial x} \right )^2 + \left (\dfrac{\partial z}{\partial y} \right ) ^2+ 1 = 4(4) +1$

$\left (\dfrac{\partial z}{\partial x} \right )^2 + \left (\dfrac{\partial z}{\partial y} \right ) ^2+ 1 = 17$

The area is,

$A = \sqrt{17} \int \int_D dA$

$A = \sqrt{17} $(Area of circle with radius 2)

$A = 4 \sqrt{17} \pi$

Is this area correct(in first quadrant). If this is wrong kindly show me the direction where I am going worng.

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  • $\begingroup$ Are you finding the area of the entire solid? Or just the face that belongs to the paraboloid? Also, if $z=2$, then you should have $x^2+y^2=\color{red}{2}$, not $4$. $\endgroup$ – user170231 Nov 18 '20 at 21:56
  • $\begingroup$ I edited the question to reflect z=4. I want to calculate the part of the paraboloid that lies under the plane z=4 $\endgroup$ – Aruha Nov 18 '20 at 22:22
  • $\begingroup$ Okay, I've adjusted the integral in my answer accordingly $\endgroup$ – user170231 Nov 18 '20 at 22:59
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If $D$ denotes the surface of the given solid, then the area of $D$ over the paraboloid is

$$\begin{align} \iint_S \sqrt{1+\left(\frac{\partial z}{\partial x}\right)^2+\left(\frac{\partial z}{\partial y}\right)^2}\,\mathrm dA&=\iint_S\sqrt{1+4(x^2+y^2)}\,\mathrm dx\,\mathrm dy\\[1ex] &=\int_0^{\frac\pi2}\int_0^2r\sqrt{1+4r^2}\,\mathrm dr\,\mathrm d\theta \end{align}$$

where $x=r\cos\theta$ and $y=r\sin\theta$. You would then get an area of $\dfrac{\left(17^{\frac32}-1\right)\pi}{24}$.

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  • $\begingroup$ Thank you. This helps! $\endgroup$ – Aruha Nov 18 '20 at 22:24

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