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Find the area of the surface z= \frac{2}{3} (x^{\frac {3}{2}}+y^{\frac{3}{2}}), 0 \leq x \leq1, 0...

Question:

Find the area of the surface {eq}z= \frac{2}{3} (x^{\frac {3}{2}}+y^{\frac{3}{2}}) {/eq}, {eq}0 \leq x \leq1, 0 \leq y \leq 1 {/eq}

Surface Area:

We can use a double integral to find the surface area of a surface. When we can write the surface as {eq}z = f (x,y) {/eq}, then we can use the following integral to find its surface area above a region in the {eq}xy {/eq}-plane {eq}D {/eq}:

{eq}\begin{align*} S &= \iint_D \sqrt{\left (\frac{\partial z}{\partial x} \right )^2 + \left (\frac{\partial z}{\partial y} \right )^2+1}\ dA \end{align*} {/eq}

Answer and Explanation:

We need the partials of {eq}z {/eq}. We have

{eq}\begin{align*} \frac{\partial z}{\partial x} &= \frac{\partial }{\partial x} \left( \frac{2}{3} (x^{3/2}+y^{3/2}) \right) \\ &= x^{1/2} \end{align*} {/eq}

and

{eq}\begin{align*} \frac{\partial z}{\partial y} &= \frac{\partial }{\partial y} \left( \frac{2}{3} (x^{3/2}+y^{3/2}) \right) \\ &= y^{1/2} \end{align*} {/eq}

And then we have

{eq}\begin{align*} \sqrt{\left (\frac{\partial z}{\partial x} \right )^2 + \left (\frac{\partial z}{\partial y} \right )^2+1} &= \sqrt{\left (x^{1/2}\right )^2 + \left (y^{1/2} \right )^2+1} \\ &= \sqrt{x+y+1} \end{align*} {/eq}

And so the area of the surface above the square {eq}D = [0,1] \times [0,1] {/eq} is

{eq}\begin{align*} S &= \int_0^1 \int_0^1 \sqrt{x+y+1}\ dy\ dx \\ &= \int_0^1 \left [ \frac23(x+y+1)^{3/2} \right ]_0^1\ dx \\ &= \frac23 \int_0^1 (x+2)^{3/2} - (x+1)^{3/2}\ dx \\ &= \frac23 \left [ \frac25(x+2)^{5/2} - \frac25(x+1)^{5/2} \right ]_0^1 \\ &= \frac4{15} \left [ (3)^{5/2} - (2)^{5/2} - (2)^{5/2} + (1)^{5/2} \right ] \\ &= \frac4{15} \left( 9\sqrt3 - 8\sqrt2+1 \right) \\ &\approx 1.4066 \end{align*} {/eq}


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Double Integrals: Applications & Examples

from AP Calculus AB & BC: Help and Review

Chapter 12 / Lesson 14
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