The base of a solid is the unit circle x^2 + y^2 = 4, and its cross-sections perpendicular to the x-axis
are rectangles of height 10. Find its volume.
Here's my work:
A for rectangle=lw
A=10*sq(4-x)
V= the integral from -4 to 4 of sq(4-x^2)*10dx
But that gets me 0, so I know I am wrong. Please help!
To find the volume of the solid, we need to consider the cross-sections perpendicular to the x-axis. In this case, the cross-sections are rectangles with a height of 10 units.
Let's analyze the problem step by step:
1. Start by visualizing the shape: The base of the solid is the unit circle centered at the origin with a radius of 2. We need to find the volume of the solid that extends perpendicular to the x-axis.
2. Determine the limits of integration: Since the base is the unit circle, it has a radius of 2. Therefore, the limits of integration for x are -2 and 2 because the circle is symmetric.
3. Express the area of the cross-section: The area of each cross-section is given by the product of the height (10 units) and the width. The width can be determined from the x-coordinate of each cross-section. Since the base is a unit circle, the width at any given x-coordinate is given by 2 times the square root of (4 - x^2) due to the Pythagorean theorem.
4. Set up the integral: The integral represents the summation of all the cross-sectional areas along the x-axis, which gives us the volume. The integral setup is as follows:
V = ∫[from -2 to 2] (10 * 2√(4 - x^2)) dx
However, there was an error in your initial attempt. You forgot to multiply by 2 when calculating the width of the rectangle. The corrected integral would be:
V = ∫[from -2 to 2] (10 * 2 * √(4 - x^2)) dx
5. Calculate the volume: Evaluate the above integral to find the volume of the solid. You can use integral techniques, such as substitution or trigonometric substitution, to evaluate this integral.
By evaluating the integral, you should obtain the correct volume of the solid.
Note: If you encounter any issues during the calculation or need further assistance, please provide the values you obtained while solving the integral, and I will be happy to help you resolve any problems.
To find the volume of the solid, we need to integrate the area of the cross-sections with respect to x.
Given that the cross-sections are rectangles with a height of 10, the width of each rectangle will be determined by the distance between the y-values of the circle at each x-value. This can be obtained by taking the square root of the radius squared minus the x-coordinate squared.
Let's go step by step to solve this:
1. Define the area of the cross-section:
A = length * width = 10 * √(4 - x^2).
2. Determine the limits of integration:
Since the base of the solid is the unit circle, the range of x-values that intersects the circle is from -2 to 2.
Therefore, we will integrate from -2 to 2.
3. Set up the integral for the volume using the area formula:
V = ∫[from -2 to 2] A dx.
V = ∫[from -2 to 2] 10 * √(4 - x^2) dx.
4. Evaluate the integral:
To evaluate the integral, we can use trigonometric substitution. Let x = 2sin(t), dx = 2cos(t) dt.
When x = -2, t = -π/2, and when x = 2, t = π/2.
Substituting these values in:
V = ∫[-π/2 to π/2] 10 * √(4 - (2sin(t))^2) * 2cos(t) dt.
V = 20 ∫[-π/2 to π/2] √(4 - 4sin^2(t)) cos(t) dt.
V = 20 ∫[-π/2 to π/2] √(4cos^2(t)) cos(t) dt. (using sin^2(t) + cos^2(t) = 1)
V = 20 ∫[-π/2 to π/2] 2cos^2(t) dt.
Now, we can use the identity cos^2(t) = (1 + cos(2t))/2:
V = 40 ∫[-π/2 to π/2] (1 + cos(2t))/2 dt.
V = (20/2)[t + (1/2)sin(2t)] [-π/2 to π/2].
V = (20/2)[(π/2 + (1/2)sin(π)) - (-π/2 + (1/2)sin(-π))].
V = 10[(π/2 + (1/2)sin(π)) + (π/2 - (1/2)sin(π))].
V = 10(π/2 + (π/2)).
V = 10π.
Thus, the volume of the solid is 10π cubic units.