Find the area of the surface. The surface z=23(x32+y32),0≤x≤1

It surface S has equation of the form z=f(x,y), where (x,y)∈D And if f has continuous partial derivatives. A(S)=∫∫D1+(dzdx)2+(dzdy)2dA Given that z=23(x32+y32) We have dzdx=x12, dzdy=y12 Therefore, Area =∫∫D1+(x12)2+(y12)2 =∫∫D1+x+ydA Given that 0≤x≤1 and 0≤y≤1 =∫01∫011+x+ydxdy =∫01(∫01(1+x+y)12dx)dy =∫01[23(1+x+y)]01dy =∫01(23(1+1+y)32)−(23(1+0+y)32)dy =23∫01(2+y)32−(1+y)32dy =23[25](2+y)

Resolved: "Find the area of the surface. The surface..."

avissidep

Answered question

2021-10-27

Find the area of the surface. The surface

z = \frac{2}{3} (\frac{x^{3}}{2} + \frac{y^{3}}{2}), 0 \leq x \leq 1

Answer & Explanation

2k1enyvp

Skilled2021-10-28Added 94 answers

It surface S has equation of the form z=f(x,y), where

(x, y) \in D

And if f has continuous partial derivatives.

A (S) = \int \int_{D} \sqrt{1 + {(\frac{d z}{d x})}^{2} + {(\frac{d z}{d y})}^{2}} d A

Given that

z = \frac{2}{3} (x^{\frac{3}{2}} + y^{\frac{3}{2}})

We have

\frac{d z}{d x} = x^{\frac{1}{2}}, \frac{d z}{d y} = y^{\frac{1}{2}}

Therefore,
Area

= \int \int_{D} \sqrt{1 + {(x^{\frac{1}{2}})}^{2} + {(y^{\frac{1}{2}})}^{2}}

= \int \int_{D} \sqrt{1 + x + y} d A

Given that

0 \leq x \leq 1

and

0 \leq y \leq 1

= \int_{0}^{1} \int_{0}^{1} \sqrt{1 + x + y} d x d y

= \int_{0}^{1} (\int_{0}^{1} {(1 + x + y)}^{\frac{1}{2}} d x) d y

= \int_{0}^{1} {[\frac{2}{3} (1 + x + y)]}_{0}^{1} d y

= \int_{0}^{1} (\frac{2}{3} {(1 + 1 + y)}^{\frac{3}{2}}) - (\frac{2}{3} {(1 + 0 + y)}^{\frac{3}{2}}) d y

= \frac{2}{3} \int_{0}^{1} {(2 + y)}^{\frac{3}{2}} - {(1 + y)}^{\frac{3}{2}} d y

= \frac{2}{3} [\frac{2}{5}] {(2 + y)}^{}

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