Question

Find the volume of the following solids using the method of your choice. The solid formed when the region bounded by \(y=x^{2}\) and \(y=2-x^{2}\) is revolved about the \(x\) -axis

Short answer

Answer: The volume of the solid is \(\frac{26\pi}{3}\).

Step-by-step solution

Determine the limits of integration.

Find the points of intersection between the two curves (i.e., where \(y=x^2\) and \(y=2-x^2\)) by solving for \(x\): \(x^2 = 2 - x^2\) \(2x^2 = 2\) \(x^2 = 1\) So, \(x = -1\) and \(x = 1\). These give us our limits of integration. The limits of integration are \(x = -1\) and \(x = 1\).

Set up the integral.

We will use the washer method to find the volume. The volume of the washer-shaped infinitesimal piece is the difference in the areas of the two circular cross-sections times the width of the piece, which is \(dx\). The area of a circle is given by \(A=\pi r^2\), where r is the radius. So, we need the radius of the two circles that are defined by the two curves, \(y=x^2\) and \(y=2-x^2\). For the outer radius (the distance from the x-axis to the curve \(y=2-x^2\)), we have \(R(x) = 2-x^2\). For the inner radius (the distance from the x-axis to the curve \(y=x^2\)), we have \(r(x) = x^2\). Now, considering the cross-sectional area of the infinitesimal washer at x, we have: \(A(x) = \pi [R(x)^2 - r(x)^2] = \pi [(2-x^2)^2 - (x^2)^2]\) Now, to find the volume, we integrate \(A(x)\) with respect to \(x\) from \(x = -1\) to \(x = 1\). \(V = \int_{-1}^1 \pi [(2-x^2)^2 - (x^2)^2] dx\)

Evaluate the integral.

We evaluate the integral to find the volume: \(V = \int_{-1}^1 \pi [(2-x^2)^2 - (x^2)^2] dx\) \(V = \pi \int_{-1}^1 [4 - 4x^2 + x^4 - x^4] dx\) \(V = \pi \int_{-1}^1 [4-4x^2] dx\) Now, we find the antiderivative and evaluate it at the limits: \(V = \pi\left[ 4x - \frac{4}{3}x^3 \right]_{-1}^1\) \(V = \pi [(4(1) - \frac{4}{3}(1)^3) - (4(-1) - \frac{4}{3}(-1)^3)]\) \(V = \pi [4-\frac{4}{3} - 4 - \frac{4}{3}]\) \(V = \pi [2 - 4 - \frac{4}{3}]\) \(V = \pi [\frac{26}{3}]\) The volume of the solid formed when the region bounded by \(y=x^2\) and \(y=2-x^2\) is revolved about the \(x\)-axis is \(\frac{26\pi}{3}\).