Volume by Revolution About a Horizontal Line

Question

Let $$R$$ be the region enclosed by the graph of $$y = 4 - x^2$$ and the $$x$$-axis. Which of the following integrals represents the volume of the solid generated when $$R$$ is revolved about the horizontal line $$y = -2$$?

Accepted Answer

$$\pi \displaystyle\int_{-2}^{2} \left[(6 - x^2)^2 - 4\right]\, dx$$

Answer

$$2\pi \displaystyle\int_{-2}^{2} x(4 - x^2)\, dx$$

Answer

$$\pi \displaystyle\int_{-2}^{2} (6 - x^2)\, dx$$

Answer

$$\pi \displaystyle\int_{-2}^{2} (4 - x^2)^2\, dx$$

Volume by Revolution About a Horizontal Line

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