# Find the area of the lateral surface over the curve C in the xy-plane and under the surface ...

## Question:

Find the area of the lateral surface over the curve C in the xy-plane and under the surface {eq}z=f(x,y) {/eq} if {eq}f(x,y) = x + y {/eq} and {eq}C : x^2 + y^2 = 4 {/eq} from (2,0) to (0,2).

## Parametric Equations:

A curve in the plane is parametrized if the set of coordinates on the curves, where {eq}\left ( x, y \right ) {/eq}

represented as the functions of a variable {eq}t {/eq}.

Let {eq}x = R \cos t , y = R \sin t {/eq}.

The formula of the area is: {eq}A = \int _{C} f \left ( x, y \right ) ds {/eq}

Given: {eq}z = f \left ( x, y \right ) = x + y {/eq}

{eq}C: x^{2} + y^{2} = 4 {/eq}

From {eq}\left ( 2,0 \right )\, to\, \left ( 0, 2 \right ) {/eq}

Let {eq}x = R \cos t , y = R \sin t {/eq}

{eq}x = 2 \cos t , y = 2 \sin t {/eq}

{eq}r \left ( t \right ) = \left ( 2 \cos t , 2 \sin t \right ) {/eq} for {eq}t \epsilon \left [ 0, \frac{\pi }{2} \right ] {/eq}

Take derivatives.

{eq}\frac{dx}{dt} = \left ( - 2 \sin t \right ) dt {/eq}

{eq}\frac{dy}{dt} = \left ( 2 \cos t \right ) dt {/eq}

Find the area.

{eq}A = \int _{C} f \left ( x, y \right ) ds {/eq}

A = \int_{0}^{\frac{\pi }{2} \left ( x + y \right ) \sqrt{\left ( \frac{dx}{dt} \right )^{2} + \left ( \frac{dy}{dt} \right )^{2} dt

{eq}= \int_{0}^{\frac{\pi }{2}} \left ( 2 \cos t + 2 \sin t \right ) \sqrt{\left ( - 2 \sin t \right )^{2} + \left ( 2 \cos t \right )^{2}} dt {/eq}

{eq}= 4 \int_{0}^{\frac{\pi }{2}} \left ( \cos t + \sin t \right ) dt {/eq}

{eq}= 4 \left ( sint - cos t \right )_{0}^{\frac{\pi }{2}} {/eq}

{eq}= 4 \left ( 1 - 0 - 0 + 1 \right ) {/eq}

{eq}= 8 {/eq}

Hence, the area is 8.