Let R be the region bounded by the following curves. Use the shell method to find the volume of...

Question:

Let {eq}R {/eq} be the region bounded by the following curves. Use the shell method to find the volume of the solid generated when {eq}R {/eq} is revolved about the {eq}x {/eq}-axis.

{eq}y = \sqrt{x}, \; y = 0, {/eq} and {eq}\; x = 4 {/eq}

Volume:

We will first find the limits of integration by solving the equations. Then we will find the radius and the height of the shell. Then we will apply the formula for the volume. The given function is a square root function.

Answer and Explanation:

{eq}x=y^2\\ y^2=0\\ y=0\\ y=\sqrt{x}\\ x=4\\ y=2\\ \text{Are the limits of integration:}\\ \text{We will find the radius andthe height of the shell:}\\ r=y\\ h=4-y^2\\ \text{The formula for the volume:}\\ V=2\pi\int_{a}^{b}rhdx\\ \text{Applying here:}\\ V=2\pi\int_{0}^{2}y(4-y^2)dy\\ =2\pi\int_{0}^{2}(4y-y^3)dy\\ \text{Applying formulas:}\\ =2\pi \left [y^2-\frac{y^4}{4} \right ]_{0}^{2}\\ =8\pi\\ \text{is the volume.} {/eq}


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