calculus question???

Question

what is the surface area of paraboloid S by revolving graph of y=x2 between x=0 and
about y-axis?

Answer 1

As mentioned before, we need to know between x=0 and x=?.

However, the general formula for the surface area is the

double integral (|Ru x Rv|)dudv, where Ru is the partial derivative with respect to u, Rv the partial derivative with respect to v. x signifies the cross-product and || the magnitude.

So, to solve your problem, we need parametric equations for the paraboloid. Seeing that it is formed by revolutions about the y-axis, you can use:

x= y^(1/2)*cos(theta), y=y, and z = y^(1/2)*sin(theta).

The partial derivative with respect to y is
<(1/2)*y^(-1/2)cos(theta),1, (1/2)*y^(-1/2)sin(theta)>

The partial derivative with respect to theta is
< -y^(1/2)*sin(theta), 0, y^(1/2)*cos(theta)>

Now, taking the cross-product Ry x Rtheta, you get


Taking the magnitude of this vector, you get
(y + 1/4)^(1/2).

Therefore, the surface area is:

....2pi.....?
integral integral (y + 1/4)^(1/2) dydtheta
.....0........0

Note: y also begins at 0 because when x is 0, y is also 0.

Answer 2

i could say structure is element of engineering which signifies that a reliable draw close on math is obligatory, pre-cal is is like algebra to the subsequent element with the Pi chart and understanding about radians and perspective measures (which seems major) yet Calculus is the learn of derivatives, (the slope of a line at a particular element) style of unnecessary no count number what container of workd your going to, in case you question me.

Answer 3

x=0 and x=?
u need to have the two values of x in order to know what the surface of the parabol is; or else, y can't be the axis

Answer 4

between x=0 and x=??

Answer 5

Ask me again at Christmas. I have to take Cal 1 this Fall. :'(

Answer 6

Do your own homework.