# Area of a region in first quadrant

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.

• 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 $4$. – user170231 Nov 18 '20 at 21:56
• I edited the question to reflect z=4. I want to calculate the part of the paraboloid that lies under the plane z=4 – Aruha Nov 18 '20 at 22:22
• Okay, I've adjusted the integral in my answer accordingly – user170231 Nov 18 '20 at 22:59

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 $$\dfrac{\left(17^{\frac32}-1\right)\pi}{24}$$.