Question

CAN SOMEONE HELP ME IN THIS INTEGRAL QUESTION PLS

''Find the surface area between the z = 1 and z = 4 planes of z = x ^ 2 + y ^ 2 paraboloid.''

Answer 1

Due to the symmetry of the paraboloid about the z-axis, you can treat this is a surface of revolution. Consider the curve
`y=x^2`, with
`1\le x\le2`, and revolve it about the y-axis. The area of the resulting surface is then

`\displaystyle2\pi\int_1^2x√(1+(y')^2)\,\mathrm dx=2\pi\int_1^2x√(1+4x^2)\,\mathrm dx=\frac{(17^(3/2)-5^(3/2))\pi}6`

But perhaps you'd like the surface integral treatment. Parameterize the surface by

`\vec s(u,v)=u\cos v\,\vec\imath+u\sin v\,\vec\jmath+u^2\,\vec k`

with
`1\le u\le2` and
`0\le v\le2\pi`, where the third component follows from

`z=x^2+y^2=(u\cos v)^2+(u\sin v)^2=u^2`

Take the normal vector to the surface to be

`(\partial\vec s)/(\partial u)*(\partial\vec s)/(\partial u)=-2u^2\cos v\,\vec\imath-2u^2\sin v\,\vec\jmath+u\,\vec k`

The precise order of the partial derivatives doesn't matter, because we're ultimately interested in the magnitude of the cross product:

`\left\|(\partial\vec s)/(\partial u)*(\partial\vec s)/(\partial v)\right\|=u√(1+4u^2)`

Then the area of the surface is

`\displaystyle\int_0^(2\pi)\int_1^2\left\|(\partial\vec s)/(\partial u)*(\partial\vec s)/(\partial v)\right\|\,\mathrm du\,\mathrm dv=\int_0^(2\pi)\int_1^2u√(1+4u^2)\,\mathrm du\,\mathrm dv`

which reduces to the integral used in the surface-of-revolution setup.