Problem 55 - Calculus Early Transcendentals

Question

Sketch a solid of revolution whose volume by the disk method is given by the following integrals. Indicate the function that generates the solid. Solutions are not unique. a. \(\int_{0}^{\pi} \pi \sin ^{2} x d x\) b. \(\int_{0}^{2} \pi\left(x^{2}+2 x+1\right) d x\)

Step-by-Step Solution

Answer

Question: Sketch the two solids of revolution whose volumes can be represented by the given integrals: a. \(\int_{0}^{\pi} \pi \sin ^{2} x d x\) and b. \(\int_{0}^{2} \pi\left(x^{2}+2 x+1\right) d x\). Answer: a. The solid of revolution formed by revolving the function \(f(x) = \sin^2{x}\) around the x-axis within the interval \([0, \pi]\) will resemble parts of two spheres, one in each quadrant. b. The solid of revolution formed by revolving the function \(f(x) = x^2 + 2x +1\) around the x-axis within the interval \([0, 2]\) will resemble a vase with a wide base and narrowing as it goes to \(x = 2\).

Step 1: Identify function and limits of integration

We are given the integral \(\int_{0}^{\pi} \pi \sin ^{2} x d x\). This suggests a function \(f(x) = \sin^2{x}\) being revolved around the x-axis within the interval \([0, \pi]\).

Step 2: Sketch graph of the function

Sketch a graph of the function \(f(x) = \sin^2{x}\) in the range \([0, \pi]\). The graph starts at the origin, reaches its maximum value at \(\frac{\pi}{2}\) and returns to zero at the end point \(\pi\).

Step 3: Sketch solid of revolution

Revolving the graph of \(f(x) = \sin^2{x}\) around the x-axis creates a solid of revolution. The shape formed will have circular cross-sections with radius equal to the height of the graph from the x-axis at each point in the interval. So, the solid will resemble parts of two spheres, one in each quadrant. b. \(\int_{0}^{2} \pi\left(x^{2}+2 x+1\right) d x\)

Step 1: Identify function and limits of integration

We are given the integral \(\int_{0}^{2} \pi\left(x^{2}+2 x+1\right) d x\). This suggests a function \(f(x) = x^2 + 2x + 1\) being revolved around the x-axis within the interval \([0, 2]\).

Step 2: Sketch graph of the function

Sketch a graph of the function \(f(x) = x^2 + 2x +1\) in the range \([0, 2]\). This function is a parabola that opens upwards and has a minimum value at \(x = -1\). In the given range, the parabola starts at \(1\) for \(x = 0\) and increases in the positive direction as \(x\) goes to \(2\).

Step 3: Sketch solid of revolution

Revolving the graph of \(f(x) = x^2 + 2x +1\) around the x-axis creates a solid of revolution. The shape formed will have circular cross-sections with radius equal to the height of the graph from the x-axis at each point in the interval. So, the solid will resemble a vase with a wide base and narrowing as it goes to \(x = 2\).