How do you find the volume of the solid obtained by rotating the region bounded by y=5x^2 ,x=1 , and y=0, about the xaxis?
Volume
graph{(y5x^2)(x1)(y)=0 [3, 3, 2, 6]}
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To find the volume of the solid obtained by rotating the region bounded by (y = 5x^2), (x = 1), and (y = 0) about the xaxis, you can use the disk method.

Determine the limits of integration by setting up the integral with respect to (x), from the lower bound to the upper bound, which are (x = 0) and (x = 1) respectively.

Find the radius of each disk at a given value of (x). For this problem, the radius, (r), is the distance from the curve (y = 5x^2) to the xaxis, which is simply (y).

Square the radius and multiply by (\pi) to find the area of each disk.

Integrate the areas of the disks from (x = 0) to (x = 1) to find the total volume using the formula for volume by disks:
[V = \int_{0}^{1} (\pi r^2) , dx]
[V = \int_{0}^{1} (\pi (5x^2)^2) , dx]
[V = \int_{0}^{1} (25\pi x^4) , dx]
[V = \pi \int_{0}^{1} (25x^4) , dx]
[V = \pi \left[\frac{25x^5}{5}\right]_{0}^{1}]
[V = \pi \left[\frac{25}{5}  \frac{0}{5}\right]]
[V = \pi \cdot 5]
[V = 5\pi]
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When evaluating a onesided 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 onesided 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 onesided 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 onesided 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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