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saw5 [17]
2 years ago
5

Two semicircles are attached to the sides of a rectangle as shown.

Mathematics
2 answers:
Tatiana [17]2 years ago
7 0
Ok so this answer is 194 try an do it you will get this answer
iragen [17]2 years ago
5 0

Answer:

194\ ft^{2}

Step-by-step explanation:

we know that

The area of the figure is equal to the area of the rectangle plus the area of one circle (Remember that the area of two semicircles is equal to the area of one circle)

step 1

Find the area of rectangle

The area of rectangle is equal to

A=bh

we have

b=8\ ft

h=18\ ft

substitute

A=8*18=144\ ft^{2}

step 2

Find the area of one circle

The area of the circle is equal to

A=\pi r^{2}

we have

r=8/2=4\ ft

\pi=3.14

substitute

A=(3.14)(4^{2})=50.24\ ft^{2}

step 3

Find the area of the figure

144\ ft^{2}+50.24\ ft^{2}=194.24\ ft^{2}

Round to the nearest whole number

194.24\ ft^{2}=194\ ft^{2}

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Answer:

The volume of the ball with the drilled hole is:

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Step-by-step explanation:

See attached a sketch of the region that is revolved about the y-axis to produce the upper half of the ball. Notice the function y is the equation of a circle centered at the origin with radius 15:

x^2+y^2=15^2\to y=\sqrt{225-x^2}

Then we set the integral for the volume by using shell method:

\displaystyle\int_5^{15}2\pi x\sqrt{225-x^2}dx

That can be solved by substitution:

u=225-x^2\to du=-2xdx

The limits of integration also change:

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\displaystyle -\int_{200}^{0}\pi \sqrt{u}du

If we flip the limits we also get rid of the minus in front, and writing the root as an exponent we get:

\displaystyle \int_{0}^{200}\pi u^{1/2}du

Then applying the basic rule we get:

\displaystyle\frac{2\pi}{3}u^{3/2}\Bigg|_0^{200}=\frac{2\pi(200\sqrt{200})}{3}=\frac{400\pi(10)\sqrt{2}}{3}=\frac{4000\pi\sqrt{2}}{3}

Since that is just half of the solid, we multiply by 2 to get the complete volume:

\displaystyle\frac{2\cdot4000\pi\sqrt{2}}{3}

=\displaystyle\frac{8000\pi\sqrt{2}}{3}

5 0
2 years ago
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