More Volume Problems — Question 6

Question 6 – Solution

1. Determine the bounds.

In the first quadrant, $\cos x=\sin x$ at $x=\pi/4$. The specified order $\cos x\ge\sin x$ holds for $0\le x\le\pi/4$.

2. Identify the triangle leg and area.

The segment in the base is a leg, not the hypotenuse. Both perpendicular legs have length$$\ell(x)=\cos x-\sin x,\qquad A(x)=\frac12\ell(x)^2.$$

3. Set up the volume and simplify.

$$V=\frac12\int_0^{\pi/4}(\cos x-\sin x)^2dx.$$$$(\cos x-\sin x)^2=\cos^2x+\sin^2x-2\sin x\cos x=1-\sin(2x).$$

4. Integrate each term.

$$V=\frac12\int_0^{\pi/4}(1-\sin(2x))dx=\frac12\left[x+\frac12\cos(2x)\right]_0^{\pi/4}.$$

5. Apply the bounds.

$$V=\frac12\left[\left(\frac\pi4+\frac12\cos\frac\pi2\right)-\left(0+\frac12\cos0\right)\right]=\boxed{\frac\pi8-\frac14}.$$

All volumes are in cubic units.