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Let $R$ be the region bounded by the curves $y = x^2$ , $x = 0$ , and $y = 6$ . What is the volume of the solid generated by rotating $R$ about the $y$ -axis?

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Find values of derivatives using limits

Full solution

Q. Let

R

be the region bounded by the curves

y = x^2

,

x = 0

, and

y = 6

. What is the volume of the solid generated by rotating

R

about the

y

-axis?

Identify Volume Calculation Method: To find the volume of the solid generated by rotating the region $R$ about the $y$ -axis, we can use the disk method. The volume $V$ of the solid is given by the integral from $y = 0$ to $y = 6$ of the area of the disk at each $y$ -value. The radius of each disk is the $x$ -value on the curve $y = x^2$ , which is $x = \sqrt{y}$ . The area of each disk is $\pi$ times the square of the radius.
Set Up Integral: Set up the integral for the volume $V$ using the disk method. The integral will be from $y = 0$ to $y = 6$ of $\pi(\sqrt{y})^2 \, dy$ .
Simplify Integrand: Simplify the integrand. Since $(\sqrt{y})^2$ is just $y$ , the integral becomes $\int_{0}^{6} \pi y \, dy$ .
Evaluate Integral: Evaluate the integral. The antiderivative of $\pi y$ with respect to $y$ is $(\pi/2)y^2$ . We need to evaluate this from $y = 0$ to $y = 6$ .
Plug in Limits: Plug in the limits of integration. The volume $V$ is $\left[\left(\frac{\pi}{2}\right)6^2\right] - \left[\left(\frac{\pi}{2}\right)0^2\right]$ .
Calculate Volume: Calculate the volume. $V = (\pi/2) \times 36 - 0 = 18\pi$ .

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