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ella [17]
3 years ago
6

find the surface area of the composite solid round your answer to the nearest tenth and explain your answer

Mathematics
2 answers:
OlgaM077 [116]3 years ago
7 0
I did most of it for you-
Volume of the cone is 13.1
Volume of the cylinder is 18.9
Now, you just subtract two of the flat circular surfaces, because the two shapes are attached, therefore those parts are not surface area. And there is your answer.
aleksandrvk [35]3 years ago
4 0
The surface area is calculated based on the surfaces which are visible from the outside. In this case, you only need to calculate the surface of lateral side on the cone, the surface of lateral side on the cylinder, and the circle on the lower base of cylinder.

Write the expression to find the surface area
sa = lateral side of cone + lateral side of cylinder + circle
sa = (π × r × s) + (2 × π × r × t) + (π × r²)

Input the numbers
sa = (π × 1 × 3) + (2 × π × 1 × 2) + (π × 1²)
sa = 3π + 4π + π
sa = 8π
sa = 8 × 3.14
sa = 25.12

to the nearest tenth
sa = 25.1

The surface area of the composite solid round is 25.1 m²
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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) &;

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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

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= \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]

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\\ \rm\longmapsto x=1.2min

Put it in eq(1)

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