Question

A solid lies between planes perpendicular to the​ y-axis at yequals0 and yequals2. The​ cross-sections perpendicular to the​ y-axis are circular disks with diameters running from the​ y-axis to the parabola x equals StartRoot 6 EndRoot y squared. Find the volume of the solid.

Answer 1

Answer:

The volume of the solid is $\frac{48\pi}{5}$

Step-by-step explanation:

Consider the provided information.

The​ cross-sections perpendicular to the​ y-axis are circular disks with diameters running from the​ y-axis to the parabola $x=\sqrt6y^2$

Therefore, diameter is $d=\sqrt6y^2$

Radius will be $r=\frac{\sqrt6y^2}{2}$

We can calculate the area of circular disk as: πr²

Substitute the respective values we get:

$A=\pi(\frac{\sqrt6y^2}{2})^2$

$A=\pi(\frac{6y^4}{4})=\frac{3\pi y^4}{2}$

Thus the volume of the solid is:

$V=\int\limits^2_0 {\frac{3\pi y^4}{2}} \, dy$

$V=[{\frac{3\pi y^5}{2\times 5}}]^2_0$

$V=\frac{48\pi}{5}$

Hence, the volume of the solid is $\frac{48\pi}{5}$

Answer 2

The volume of solid represent the how much space an object occupied. In the given problem volume can be determine by taking the integration of Area of solid.

The volume of solid is $\frac{48\pi }{5}$.

Given:

The​ cross-sections perpendicular to the​ y-axis are circular disks with diameters running from the​ y-axis to the parabola is $x=\sqrt{6}y^2$.

The diameter of the solid is $d=\sqrt{6}y^2$.

Calculate the radius of the solid.

$r=\frac{d}{2}\\r=\frac{\sqrt{6}y^2}{2}$

Write the expression for area of circular disk.

$A=\pi r^2\\A=\pi (\frac{\sqrt{6}y^2}{2})^2\\A=\frac{3\pi y^4}{2}$

Calculate the volume of solid.

$V=\int\limits^2_0 {\frac{3\pi y^4 }{2} } \, dy\\V=[\frac{3\pi y^5}{2\times 5}]_{0}^{2}\\V=\frac{48\pi }{5}$

Thus, the volume of solid is $\frac{48\pi }{5}$ .

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