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GrogVix [38]
2 years ago
12

Find the area of the surface. The part of the surface z = xy that lies within the cylinder x2 + y2 = 36.

Mathematics
1 answer:
rewona [7]2 years ago
3 0

Answer:

Step-by-step explanation:

From the given information:

The domain D of integration in polar coordinates can be represented by:

D = {(r,θ)| 0 ≤ r ≤ 6, 0 ≤ θ ≤ 2π) &;

The partial derivates for z = xy can be expressed  as:

y =\dfrac{\partial z}{\partial x} , x = \dfrac{\partial z}{\partial y}

Thus, the area of the surface is as follows:

\iint_D \sqrt{(\dfrac{\partial z}{\partial x})^2+ (\dfrac{\partial z}{\partial y})^2 +1 }\ dA = \iint_D \sqrt{(y)^2+(x)^2+1 } \ dA

= \iint_D \sqrt{x^2 +y^2 +1 } \ dA

= \int^{2 \pi}_{0} \int^{6}_{0} \ r  \sqrt{r^2 +1 } \ dr \ d \theta

=2 \pi \int^{6}_{0} \ r  \sqrt{r^2 +1 } \ dr

= 2 \pi \begin {bmatrix} \dfrac{1}{3}(r^2 +1) ^{^\dfrac{3}{2}} \end {bmatrix}^6_0

= 2 \pi \times \dfrac{1}{3}  \Bigg [ (37)^{3/2} - 1 \Bigg]

= \dfrac{2 \pi}{3} \Bigg [37 \sqrt{37} -1 \Bigg ]

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