Application of Integral - Rotation of regions

[Vertciel]

Hello everyone:

I am having trouble with the following problem on finding the new volume of a region after rotation. The question does not specify which integration method to use, so I chose to use the Shell Method.

Thank you.

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1. Find the volume generated by rotating the region bounded by the curves \(\displaystyle f(x) = x^2 - 2\) and \(\displaystyle g(x) = 7\) about the line \(\displaystyle y = 7\).

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To find the limits of integration:

\(\displaystyle x^2 - 2 = 7\)

\(\displaystyle x = -3, 3\)

The Shell Method: \(\displaystyle V = 2 \pi \int_a^{b} d(y)h(y) dy\)

Since:

\(\displaystyle d(y) = y\),

\(\displaystyle h(y) = -y^2 + 9\).

Therefore: \(\displaystyle V = 2 \pi \int_{-3}^{3} y(-y^2 + 9) dy\)

\(\displaystyle V = 2 \pi \int_{-3}^{3} (-y^3 + 9y) dy\)

When I integrate this, I get an answer of 0 which is not correct as there must be volume in the region. Could anyone please tell me where I may have erred?

 

[galactus]

You can use disks and it's not too bad.

You are using x limits after you put it in terms of y. The y limits should be -2 to 7.

\(\displaystyle {\pi}\int_{-3}^{3}(x^{2}-9)^{2}dx\)

Shells would be like so:

\(\displaystyle 4{\pi}\int_{-2}^{7}(7-y)\sqrt{y+2}dy\)

 

[Deleted member 4993]

Hello everyone:

I am having trouble with the following problem on finding the new volume of a region after rotation. The question does not specify which integration method to use, so I chose to use the Shell Method.<<< Actually disk method will give you much easier integration

Use your graphing calculator and sketch the situation.

The figure is symmetric about x=0 - so you can deal with half the figure and multiply by 2.

Then the limits are x = 0 to x = 3

and the disk radius = [7-(x^2 - 2)]

Thank you.

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1. Find the volume generated by rotating the region bounded by the curves \(\displaystyle f(x) = x^2 - 2\) and \(\displaystyle g(x) = 7\) about the line \(\displaystyle y = 7\).

---

To find the limits of integration:

\(\displaystyle x^2 - 2 = 7\)

\(\displaystyle x = -3, 3\)

The Shell Method: \(\displaystyle V = 2 \pi \int_a^{b} d(y)h(y) dy\)

Since:

\(\displaystyle d(y) = y\),

\(\displaystyle h(y) = -y^2 + 9\).

Therefore: \(\displaystyle V = 2 \pi \int_{-3}^{3} y(-y^2 + 9) dy\)

\(\displaystyle V = 2 \pi \int_{-3}^{3} (-y^3 + 9y) dy\)

When I integrate this, I get an answer of 0 which is not correct as there must be volume in the region. Could anyone please tell me where I may have erred?