# How do you find the area enclosed by #y=sin x# and the x-axis for #0≤x≤pi# and the volume of the solid of revolution, when this area is rotated about the x axis?

Area = 2 areal units.

Volume of the solid of revolution =

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To find the area enclosed by ( y = \sin x ) and the x-axis for ( 0 \leq x \leq \pi ), we integrate ( y = \sin x ) with respect to x over the given interval:

[ A = \int_{0}^{\pi} \sin x , dx ]

To find the volume of the solid of revolution when this area is rotated about the x-axis, we use the disk method. The volume ( V ) is given by:

[ V = \pi \int_{0}^{\pi} (\sin x)^2 , dx ]

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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- Calculus 2. Is this the right answer?

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