Math Problem Statement

find the volume of the solid obtained when the region under the curve y=xsqrt(4-x^2) from x=0 to x=2 is rotated about the y-axis.

Solution

To find the volume of the solid obtained by rotating the region under the curve y=x4x2y = x\sqrt{4 - x^2} from x=0x = 0 to x=2x = 2 around the y-axis, we use the method of cylindrical shells.

The formula for the volume of a solid generated by revolving a region around the y-axis is:

V=2πabxf(x)dxV = 2\pi \int_{a}^{b} x \cdot f(x) \, dx

Here, f(x)=x4x2f(x) = x\sqrt{4 - x^2}, and the limits of integration are x=0x = 0 to x=2x = 2.

Thus, the volume is given by:

V=2π02xx4x2dxV = 2\pi \int_{0}^{2} x \cdot x\sqrt{4 - x^2} \, dx

Simplify the integrand:

V=2π02x24x2dxV = 2\pi \int_{0}^{2} x^2\sqrt{4 - x^2} \, dx

Step 1: Substitution

We perform the substitution:

u=4x2du=2xdxu = 4 - x^2 \quad \Rightarrow \quad du = -2x \, dx

When x=0x = 0, u=4u = 4, and when x=2x = 2, u=0u = 0.

Rewriting the integral in terms of uu:

V=2π40(4u)u2duV = 2\pi \int_{4}^{0} \frac{(4 - u)\sqrt{u}}{-2} \, du

Flipping the limits and simplifying:

V=2π04(4u)uduV = 2\pi \int_{0}^{4} (4 - u)\sqrt{u} \, du

Step 2: Distribute and Integrate

Now, distribute u\sqrt{u} to each term:

V=2π04(4uuu)duV = 2\pi \int_{0}^{4} \left( 4\sqrt{u} - u\sqrt{u} \right) \, du

This becomes:

V=2π(404u1/2du04u3/2du)V = 2\pi \left( 4 \int_{0}^{4} u^{1/2} \, du - \int_{0}^{4} u^{3/2} \, du \right)

Now, integrate each term:

u1/2du=23u3/2,u3/2du=25u5/2\int u^{1/2} \, du = \frac{2}{3} u^{3/2}, \quad \int u^{3/2} \, du = \frac{2}{5} u^{5/2}

Step 3: Evaluate the Integrals

Now substitute back the limits of integration:

404u1/2du=42343/2=4238=6434 \int_{0}^{4} u^{1/2} \, du = 4 \cdot \frac{2}{3} \cdot 4^{3/2} = 4 \cdot \frac{2}{3} \cdot 8 = \frac{64}{3}

04u3/2du=2545/2=2532=645\int_{0}^{4} u^{3/2} \, du = \frac{2}{5} \cdot 4^{5/2} = \frac{2}{5} \cdot 32 = \frac{64}{5}

Step 4: Combine the Results

Now subtract the integrals:

V=2π(643645)V = 2\pi \left( \frac{64}{3} - \frac{64}{5} \right)

To combine the terms, get a common denominator:

V=2π(3201519215)=2π12815V = 2\pi \left( \frac{320}{15} - \frac{192}{15} \right) = 2\pi \cdot \frac{128}{15}

Finally:

V=256π15V = \frac{256\pi}{15}

Thus, the volume of the solid is 256π15\frac{256\pi}{15}.


Do you want further details on the steps? Here are 5 related questions:

  1. How would you calculate the volume if the region was rotated around the x-axis?
  2. Can this method be used for curves that intersect the axis of rotation?
  3. How does the cylindrical shell method compare to the disk method?
  4. What is the geometric significance of the substitution used in this problem?
  5. Can the integral limits change if the curve is shifted vertically?

Tip: When rotating regions, always visualize the axis of rotation to choose the right method: shells (for rotation around a vertical axis) or disks (for rotation around a horizontal axis).

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Math Problem Analysis

Mathematical Concepts

Cylindrical Shell Method
Definite Integration
Substitution Method
Solid of Revolution

Formulas

Volume of a solid using cylindrical shells: V = 2π ∫[a,b] x * f(x) dx
Substitution: u = 4 - x^2, du = -2x dx
Integrals of powers of u: ∫u^(1/2) du = (2/3) u^(3/2), ∫u^(3/2) du = (2/5) u^(5/2)

Theorems

Fundamental Theorem of Calculus
Substitution in Definite Integrals

Suitable Grade Level

College Level - Calculus II