Question

The region above the curve y = x3, under the line y =1, and between x=0 and x = 1; about the x-axis. find volume?

Answer 1

For a better understanding of the question please go through the diagram in the file that has been attached. It has been attempted to make the diagram as much self explanatory as possible.

Let us begin with the solution.

We need the volume of the shaded area. Please note that an equivalent area is to be shaded below the x Axis.

Now, we know that the equation given to us is $y=x^3$, so, when x=0, y=0 and when x=1, y=1 too.

Thus, the limits of integration will be from x=0 to x=1.

The required volume, V can thus be found as:

$V=\pi\int_{0}^{1}[(1)^2-(x^3)^2])dx=\pi\int_{0}^{1}[1-x^6]dx$

$\therefore V=\pi[[x_{0}^{1}]-[\frac{x^7}{7}]_{0}^{1}]=\pi[1-\frac{1}{7}]=\frac{6\pi}{7}$

Thus the required volume is $\frac{6\pi}{7}$

Answer 2

Step-by-step explanation:

For a better understanding of the question please go through the diagram in the file that has been attached. It has been attempted to make the diagram as much self explanatory as possible.

Let us begin with the solution.

We need the volume of the shaded area. Please note that an equivalent area is to be shaded below the x Axis.

Now, we know that the equation given to us is y=x^3y=x

3

, so, when x=0, y=0 and when x=1, y=1 too.

Thus, the limits of integration will be from x=0 to x=1.

The required volume, V