Math Problem Statement

Find the volume of the solid generated by revolving the region bounded by y equals 3 StartRoot Bold sin x EndRoot​, y equals 0 comma and x 1 equals StartFraction pi Over 4 EndFraction and x 2 equals StartFraction 3 pi Over 4 EndFraction about the​ x-axis. Question content area bottom Part 1 The volume of the solid is    enter your response here cubic units. ​(Type an exact​ answer, using pi as​ needed.)

Solution

To find the volume of the solid generated by revolving the region bounded by y=3sin(x)y = 3\sqrt{\sin(x)}, y=0y = 0, x1=π4x_1 = \frac{\pi}{4}, and x2=3π4x_2 = \frac{3\pi}{4} around the x-axis, we can apply the disk method.

Disk Method Formula:

V=πx1x2[f(x)]2dxV = \pi \int_{x_1}^{x_2} [f(x)]^2 \, dx Where f(x)=y=3sin(x)f(x) = y = 3\sqrt{\sin(x)}, and we are revolving around the x-axis from x1=π4x_1 = \frac{\pi}{4} to x2=3π4x_2 = \frac{3\pi}{4}.

Step 1: Set up the integral

The radius of each disk is given by y=3sin(x)y = 3\sqrt{\sin(x)}, so: V=ππ43π4(3sin(x))2dxV = \pi \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \left(3\sqrt{\sin(x)}\right)^2 \, dx

Step 2: Simplify the integrand

(3sin(x))2=9sin(x)\left(3\sqrt{\sin(x)}\right)^2 = 9\sin(x) Thus, the integral becomes: V=ππ43π49sin(x)dxV = \pi \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} 9\sin(x) \, dx

Step 3: Compute the integral

V=9ππ43π4sin(x)dxV = 9\pi \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \sin(x) \, dx The antiderivative of sin(x)\sin(x) is cos(x)-\cos(x), so: V=9π[cos(x)]π43π4V = 9\pi \left[-\cos(x)\right]_{\frac{\pi}{4}}^{\frac{3\pi}{4}}

Step 4: Evaluate the definite integral

First, calculate cos(x)\cos(x) at the bounds: cos(3π4)=12,cos(π4)=12\cos\left(\frac{3\pi}{4}\right) = -\frac{1}{\sqrt{2}}, \quad \cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}

Now, plug these values into the expression: V=9π[(12)(12)]V = 9\pi \left[-\left(-\frac{1}{\sqrt{2}}\right) - \left(-\frac{1}{\sqrt{2}}\right)\right] V=9π[12+12]V = 9\pi \left[\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}\right] V=9π[22]V = 9\pi \left[\frac{2}{\sqrt{2}}\right] V=9π×2V = 9\pi \times \sqrt{2}

Final Answer:

The volume of the solid is 9π29\pi\sqrt{2} cubic units.

Would you like more details on this solution, or do you have any other questions?

Here are five related questions to deepen your understanding:

  1. How does the disk method differ from the washer method in finding volumes?
  2. What would change in the integral if the region were revolved around the y-axis instead?
  3. How does the behavior of sin(x)\sin(x) affect the shape of the solid of revolution?
  4. Can you apply the cylindrical shell method to this problem? If so, how?
  5. What is the effect of changing the limits of integration on the volume?

Tip: When setting up integrals for solids of revolution, always check whether the function you're revolving requires using the disk or washer method!

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Math Problem Analysis

Mathematical Concepts

Calculus
Volume of Solids of Revolution
Disk Method
Trigonometric Functions

Formulas

V = π ∫[f(x)]² dx
Antiderivative of sin(x) = -cos(x)

Theorems

Disk Method Theorem for Volume

Suitable Grade Level

Grades 11-12, College-level Calculus