Find the surface area of the portion of the cone that is above the region in the first quadrant bounded by the line , and the parabola .

Question

Find the surface area of the portion of the cone  that is above the region in the first quadrant bounded by the line  , and the parabola  .

Quizplus · Accepted Answer

First, we find the equation of the parabola. Since it passes through the origin and has a vertical axis of symmetry, it can be written in the form $y=ax^2$. Substituting the point $(4,2)$, we get $2=16a$ or $a=\frac{1}{8}$. So the equation of the parabola is $y=\frac{1}{8}x^2$.

Next, we find the intersection point of the line and the parabola. Setting the equations equal, we get $x=8$. So the region in the first quadrant that we're interested in is bounded by the $x$-axis, the line $x=8$, and the parabola $y=\frac{1}{8}x^2$.

Now we need to find the slant height of the cone. Drawing a vertical line from the apex of the cone to the $x$-axis, we see that it forms a $30^\circ$ angle with the horizontal. So the slant height is $10\operatorname{sin}(30^\circ)=5$.

Using Pythagoras, we can find the radius of the base of the cone. We have $r^2=h^2-x^2$, where $h$ is the height of the cone and $x$ is the distance from the apex to the point on the base directly below it. At $x=8$, we have $h=10$ (since the cone has height 10), so $r^2=100-64=36$, or $r=6$.

Finally, we can find the area of the portion of the cone that we're interested in. The lateral surface area of a cone is $\pi r\ell$, where $\ell$ is the slant height. So the lateral surface area of the portion of the cone that we want is $\frac{1}{3}\pi(6^2)(5)=60\pi$. To this, we need to add the area of the base, which is $\pi(6^2)=36\pi$. Therefore, the total surface area is $96\pi$.
 
Looking at the choices, the only one that matches is D.

Find the Surface Area of the Portion of the Cone