# Solve Volume of Solid: y=x, y=x^1/2, y=1

• necessary
In summary, the conversation is about finding the volume of a solid by rotating a vertical strip around y=1. The correct integral is V = ∫0^1 π((1-x)^2 - (1-√x)^2) dx. The mistake was rotating around the wrong axis and using the wrong inner and outer radius.
necessary
volume of solid:(

question is; y=x and y=x^1/2 about y =1

ok I showed my work..

I used this formula
integral Pı(outer r)^2-Pı(inner)^2

inside integral from y=0 to y=1 pi(1-y)^2 - pi(1-y^2)^2

my result is 7/6 pi..:( where am I wrong??

integrate it with respect to x. The inner radius is $$1-\sqrt{x}$$ and the outer radius is $$1- x$$.

So you have $$\int_{0}^{1} \pi((1-x)^{2}-(1-\sqrt{x})^{2}) \; dx$$

Last edited:
Well, your integrand is utterly wrong!
You are to rotate the VERTICAL strip $\sqrt{x}\leq{y}\leq{x}$ around y=1.
That gives you the integral:
$$V=\int_{0}^{1}\pi((1-x)^{2}-(1-\sqrt{x})^{2})dx=\pi\int_{0}^{1}(x^{2}-3x+2\sqrt{x})dx$$

Apart from being wrongly signed, you have rotated your figure around the line x=1, rather than around y=1.

## 1. How do I determine the volume of a solid with the given equations?

The volume of a solid can be determined by using the formula V = ∫[a,b]A(x)dx, where A(x) represents the cross-sectional area of the solid at a given point along the x-axis and [a,b] represents the limits of integration. In this case, A(x) would be the area between the curves y=x and y=x^1/2, from x=0 to x=1. The resulting integral would give you the volume of the solid.

## 2. Can I use a calculator to solve for the volume?

Yes, you can use a graphing calculator or a numerical integration calculator to solve for the volume. Simply input the equations into the calculator and use the appropriate function to find the integral.

## 3. What units should I use for the volume?

The units for the volume will depend on the units used for the equations. If the equations are in standard units (such as meters or feet), then the resulting volume will be in cubic units (such as cubic meters or cubic feet).

## 4. Is there a simpler way to find the volume without using calculus?

No, for these specific equations, calculus is the most efficient and accurate method for finding the volume of the solid. However, for other types of solids, there may be simpler methods available.

## 5. What does the graph of these equations look like?

The graph of these equations would show two intersecting lines (y=x and y=x^1/2) and a line at y=1. The solid would be the region bounded by these lines and the x-axis, which would have a triangular cross-section at the base and a straight edge at the top.

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