1. **Problem:** Find the volume of the solid of revolution formed by revolving the region bounded by $y = x^2$ and $x = y^2$ about the line $y = 1$. 2. **Identify the curves and region:** - The curves are $y = x^2$ and $x = y^2$. - Rewrite $x = y^2$ as $y = \sqrt{x}$ (considering positive root since $y \geq 0$). - The region bounded is between $y = x^2$ and $y = \sqrt{x}$. 3. **Find points of intersection:** Set $x^2 = \sqrt{x}$. $$x^2 = x^{1/2} \implies x^{3/2} = 1 \implies x = 1^{2/3} = 1$$ Also, $x=0$ is an intersection. So, the region is between $x=0$ and $x=1$. 4. **Set up the volume integral using the washer method about $y=1$:** The distance from $y=1$ to the curve is the radius of the washers. Outer radius $R = 1 - y_{lower} = 1 - x^2$ (since $y=x^2$ is lower curve) Inner radius $r = 1 - y_{upper} = 1 - \sqrt{x}$ Volume: $$V = \pi \int_0^1 \left[(1 - x^2)^2 - (1 - \sqrt{x})^2\right] dx$$ 5. **Expand and simplify the integrand:** $$(1 - x^2)^2 = 1 - 2x^2 + x^4$$ $$(1 - \sqrt{x})^2 = 1 - 2\sqrt{x} + x$$ Subtracting: $$1 - 2x^2 + x^4 - (1 - 2\sqrt{x} + x) = -2x^2 + x^4 + 2\sqrt{x} - x$$ 6. **Integral becomes:** $$V = \pi \int_0^1 (-2x^2 + x^4 + 2x^{1/2} - x) dx$$ 7. **Integrate term-by-term:** $$\int_0^1 -2x^2 dx = -2 \cdot \frac{x^3}{3} \Big|_0^1 = -\frac{2}{3}$$ $$\int_0^1 x^4 dx = \frac{x^5}{5} \Big|_0^1 = \frac{1}{5}$$ $$\int_0^1 2x^{1/2} dx = 2 \cdot \frac{2}{3} x^{3/2} \Big|_0^1 = \frac{4}{3}$$ $$\int_0^1 -x dx = -\frac{x^2}{2} \Big|_0^1 = -\frac{1}{2}$$ 8. **Sum the integrals:** $$V = \pi \left(-\frac{2}{3} + \frac{1}{5} + \frac{4}{3} - \frac{1}{2}\right) = \pi \left(\frac{-10}{15} + \frac{3}{15} + \frac{20}{15} - \frac{7.5}{15}\right)$$ $$= \pi \left(\frac{-10 + 3 + 20 - 7.5}{15}\right) = \pi \cdot \frac{5.5}{15} = \frac{11\pi}{30}$$ 9. **Final answer:** $$\boxed{\frac{11\pi}{30}}$$ This corresponds to option d.