QuestionThe answer above is NOT correct.
Find the volume of the solid whose base is the region in the first quadrant bounded by $y=x^{6}, y=1$, and the $y$-axis and whose cross-sections perpendicular to the $x$ axis are squares.

Studdy Solution
Evaluate the integral:
First, expand the integrand:
$$(1 - x^6)^2 = 1 - 2x^6 + x^{12}$$
Now, integrate term by term:
$$V = \int_{0}^{1} (1 - 2x^6 + x^{12}) \, dx$$
$$= \left[ x - \frac{2}{7}x^7 + \frac{1}{13}x^{13} \right]_{0}^{1}$$
Evaluate the definite integral:
$$V = \left( 1 - \frac{2}{7} \times 1 + \frac{1}{13} \times 1 \right) - \left( 0 - 0 + 0 \right)$$
$$= 1 - \frac{2}{7} + \frac{1}{13}$$
Combine the fractions:
$$= \frac{91}{91} - \frac{26}{91} + \frac{7}{91}$$
$$= \frac{91 - 26 + 7}{91}$$
$$= \frac{72}{91}$$
The volume of the solid is:
$$\boxed{\frac{72}{91}}$$