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3 years ago

15

Find the volume of the solid formed by revolving the region bounded by the graphs of y = x^3, x = 2, and y = 1 about the y-axis.

1 answer:

andrezito [222]3 years ago

3 0

Since the rotation is about the y-axis, I'll integrate by dy.

$\displaystyle y=x^3\\x=\sqrt[3]y\\\\V=\pi \int \limits_1^8(2^2-(\sqrt[3]y)^2)\, dy\\V=\pi \Big[4x-\dfrac{3}{5}x^{\tfrac{5}{3}}\Big]_1^8\\V=\pi \left(4\cdot8-\dfrac{3}{5}\cdot8^{\tfrac{5}{3}\right-\left(4\cdot1-\dfrac{3}{5}\cdot1^{\tfrac{5}{3}\right)\right)\\V=\pi \left(32-\dfrac{96}{5}-\left(4-\dfrac{3}{5}\right)\right)\\V=\pi \left(\dfrac{64}{5}-\dfrac{17}{5}\right)\\V=\dfrac{47\pi}{5}$

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Answer:

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Step-by-step explanation:

$(\frac{3}{4})^6 \times (\frac{16}{9})^5=(\frac{4}{3})^{x+2}$

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This implies

x+2=4

and

-(x+2)=-4.

x+2=4 implies x=2 since subtract 2 on both sides gives us x=2.

Solving -(x+2)=-4 should give us the same value.

Multiply both sides by -1:

x+2=4

It is the same equation as the other.

You will get x=2 either way.

Let's check:

$(\frac{3}{4})^6 \times (\frac{16}{9})^5=(\frac{4}{3})^{2+2}$

$(\frac{3}{4})^6 \times (\frac{16}{9})^5=(\frac{4}{3})^{4}$

Put both sides into your calculator and see if you get the same thing on both sides:

Left hand side gives 256/81.

Right hand side gives 256/81.

Both side are indeed the same for x=2.

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