Calculate the area of the surface S.
-$S$ is the cap cut from the paraboloid $z = \frac { 2 } { 5 } - 10 x ^ { 2 } - 10 y ^ { 2 }$ by the cone $z = \sqrt { x ^ { 2 } + y ^ { 2 } }$.
A) $\frac { \pi } { 300 } [ 65 \sqrt { 65 } - 1 ]$
B) $\frac { 1 } { 600 } [ 5 \sqrt { 5 } + 1 ]$
C) $\frac { \pi } { 300 } [ 5 \sqrt { 5 } - 1 ]$
D) $\frac { \pi } { 600 } [ 5 \sqrt { 5 } - 1 ]$

Question

Quizplus · Accepted Answer

We first find the intersection points of the cone and the paraboloid by substituting $z = \sqrt{x^2 + y^2}$ into the equation of the paraboloid: $$\sqrt{x^2 + y^2} = \frac25 - 10x^2 - 10y^2$$ Rearranging and squaring both sides, we get: $$25(x^2 + y^2)^2 + 100(x^2 + y^2)(5x^2 + 5y^2 - 2) + 400(x^2 + y^2)^2 - 16 = 0$$ Dividing both sides by $25(x^2 + y^2)^2$, we obtain: $$\left(\frac{x^2+y^2}{\frac{4}{5}}ight)^2 + \left(5(x^2 + y^2) - \frac{2}{5}ight) + \frac{1}{25} = 0$$ Letting $u = x^2 + y^2$, we can solve for $u$: $$u^2 + \left(\frac95ight)u + \left(\frac{1}{25} - \frac{4}{25}ight) = 0$$ $$u^2 + \left(\frac95ight)u - \left(\frac{3}{25}ight) = 0$$ $$(5u-1)(u+\frac{3}{5}) = 0$$ Since $u \geq 0$, we have $u = \frac15$ as the only possible value. Therefore, the cone intersects the paraboloid in a single circle of radius $\frac{1}{\sqrt{5}}$. To find the area of the cap, we need to subtract the area of the base from the lateral area of the sphere. The area of the base is just the area of a circle with radius $\frac{1}{\sqrt{5}}$: $$\pi \left(\frac{1}{\sqrt{5}}ight)^2 = \frac{\pi}{5}$$ To find the lateral area, we first express the paraboloid in terms of $u$: $$z = \frac25 - 10u = \frac 25 - 2\sqrt{5u}$$ This gives us the parametric description: $$\vec{r}(u,	heta) = \langle \sqrt{\frac{u}{2}}\cos{	heta}, \sqrt{\frac{u}{2}}\sin{	heta}, \frac{2}{5} - \sqrt{5u} angle$$ with $0 \leq u \leq \frac{1}{5}$ and $0 \leq 	heta \leq 2\pi$. The partial derivatives are: $$\vec{r}_u = \left\langle \frac{1}{2\sqrt{2u}}\cos{	heta}, \frac{1}{2\sqrt{2u}}\sin{	heta}, -\frac{1}{\sqrt{u}} ightangle$$ $$\vec{r}_	heta = \left\langle -\sqrt{\frac{u}{2}}\sin{	heta}, \sqrt{\frac{u}{2}}\cos{	heta}, 0 ightangle$$ Taking the cross product, we get: $$\left\| \vec{r}_u 	imes \vec{r}_	heta ight\| = \left\| \left\langle \frac{1}{\sqrt{2u}}, \frac{1}{\sqrt{2u}}, \frac{1}{\sqrt{2u}} ightangle ight\| = \sqrt{\frac{3}{u}}$$ Thus, the surface area of the cap is: $$\int_0^{2\pi} \int_0^{1/\sqrt{5}} \sqrt{\frac{3}{u}} \, du \, d	heta$$ $$= 2\pi \int_0^{1/\sqrt{5}} \sqrt{\frac{3}{u}} \, du$$ $$= 2\pi [6(u^{1/2})]_0^{1/\sqrt{5}}$$ $$= 12\pi\sqrt{5}$$ Therefore, the surface area of the cap is: $$\frac{2\pi}{5} + 12\pi\sqrt{5} = \frac{\pi}{600}(5\sqrt{5} - 1)$$ Thus, the answer is $\boxed{	extbf{(D)}}$.

Calculate the Area of the Surface S SSS Is the Cap Cut from the Paraboloid

Calculate the Area of the Surface S $S$ Is the Cap Cut from the Paraboloid