Find the area of the portion of the surface of the sphere $x^2+y^2+z^2=4x$ that is cut off by a nappe of the cone $y^2+z^2=x^2$.

I'm having trouble with a Calculus question where I am supposed to use the following formula to calculate the area of a surface:

Suppose that $$f$$ and its first-order partial derivatives are continuous in the closed region $$R$$ in the $$xy$$ plane. So, if $$S$$ is the measure of the area of the surface $$z=f(x,y)$$ above $$R$$,

$$S = \int_R\int \sqrt{f_x^2(x,y)+f_y^2(x,y)+1} dxdy$$

(Here, $$f_x(x,y)$$ and $$f_y(x,y)$$ stand for the first-order partial derivatives with respect to $$x$$ and $$y$$, respectively.)

The question asks me the following:

Find the area of the portion of the surface of the sphere $$x^2+y^2+z^2=4x$$ that is cut off by a nappe of the cone $$y^2+z^2=x^2$$.

I made the following plot of the situation described in the question. To make things simpler, I plotted only the intersection of the sphere and the cone with the $$xy$$ plane:

The image above shows the part of the sphere (indicated in cyan (light blue) color) that is cut off by a nappe of the cone. Notice that the intersection of the sphere with the $$xy$$ plane is the circle $$y^2=4x-x^2$$, and the intersection of the cone with the $$xy$$ plane are the lines $$y=x$$ and $$y=-x$$.

To apply the given formula, I have to find the region $$R$$ over which the double integral will be computed. From the image, the region of integration $$R$$ is the region between the line $$x=2$$ and the portion of the circumference marked in red. The desired portion of the surface of the sphere is above this region $$R$$.

So, the integration limits are the following:

$$\int_2^4\int_0^\sqrt{4x-x^2} \cdots dydx$$

In other words, the outer integral is computed with $$x$$ going from $$2$$ to $$4$$, and the inner integral is computed with $$y$$ going from $$0$$ to $$\sqrt{4x-x^2}$$ (notice that I have switched $$dx$$ and $$dy$$ from the original formula for convenience).

Now, if I compute the first-order partial derivatives indicated in the formula, I obtain the following results: $$f_x(x,y) = \dfrac{2-x}{\sqrt{4x-x^2-y^2}}$$ and $$f_y(x,y) = -\dfrac{y}{\sqrt{4x-x^2-y^2}}$$.

The double integral is, then:

$$\int_2^4\int_0^\sqrt{4x-x^2} \sqrt{\frac{(2-x)^2+y^2}{4x-x^2-y^2} + 1} dydx$$

which, after some simplification, becomes:

$$\int_2^4\int_0^\sqrt{4x-x^2} \sqrt{\frac{4}{4x-x^2-y^2}} dydx$$

I get stuck at this point, because I can't seem to find a way of solving this integral. I have tried to solve the inner integral

$$\int_0^\sqrt{4x-x^2} \sqrt{\frac{4}{4x-x^2-y^2}} dy$$

but without any success. Is the set up of the double integral correct? Or is there a way to solve the above integral and get a real result?

• Have you tried doing the problem in polar coordinates? Jun 18 '13 at 17:07
• @TedShifrin: yes, I tried substituting $x^2+y^2=r^2$, $x=r\cos{\theta}$ and $dydx = rdrd\theta$, but the result was messier than in Cartesian coordinates.
– favq
Jun 18 '13 at 17:12
• Because of the unexpected shape (symmetry about the $x$-axis rather than symmetry about the $z$-axis), I would recommend doing the integral in $yz$-coordinates, or, better yet, in polar coordinates in the $yz$-plane, not the $xy$-plane. Better yet, can't we do the problem with no integrals at all? Isn't this just the area of a hemisphere? :) Jun 18 '13 at 17:31
• @TedShifrin: Thank you for your suggestion, I will try that. Yes, in fact, it is just the area of a hemisphere; I had already realized that before. But I still want to get to the correct result through that formula.
– favq
Jun 18 '13 at 17:59
• Well done! Sometimes it takes a twist on the usual approach :) Jun 19 '13 at 20:24

I will solve this problem again using the suggestion by Ted Shifrin in the comments to this question above. Doing the integral in polar coordinates, the solution becomes much simpler.

I will take the polar plane to be the $yz$ plane.

In $yz$ coordinates, the desired surface is the function $x=f(y,z)=\sqrt{4-y^2-z^2}+2$ (which corresponds to the right hemisphere of the sphere represented in the picture in the question).

As required by the formula, the surface $x=f(y,z)=\sqrt{4-y^2-z^2}+2$ is above a region $R$ which lies on the $yz$ plane (polar plane). In this case, the region of integration $R$ is the projection of the sphere on the polar plane; in other words, $R$ is a circle of radius 2 centered at $(x,y,z)=(2,0,0)$.

Thus, the limits of integration in polar coordinates will be:

$$\int_0^2\int_0^{2\pi} \cdots r d\theta dr$$

The partial derivatives of $f(x,y,z)$ are $f_y = -\dfrac{y}{\sqrt{4-y^2-z^2}}$ and $f_z = -\dfrac{z}{\sqrt{4-y^2-z^2}}$.

The square root inside the double integral is:

$$\sqrt{f_y^2(y,z)+f_z^2(y,z)+1} = \sqrt{ \dfrac{y^2}{4-y^2-z^2} + \dfrac{z^2}{4-y^2-z^2} + 1 }$$

$$= \sqrt{ \dfrac{4}{4-y^2-z^2} }$$

Converting the above expression to polar coordinates (substituting $y^2+z^2=r^2$):

$$\sqrt{ \dfrac{4}{4-r^2} }$$

Thus, the double integral in polar coordinates is:

$$\int_0^2\int_0^{2\pi} \sqrt{ \dfrac{4}{4-r^2} } r d\theta dr$$

Solving the inner integral, the above expression reduces to $\int_0^2 4 \pi r \sqrt{ \dfrac{1}{4-r^2} } dr$, which can be solved by substitution (for example, by setting $u=4-r^2$ and $du=-2rdr$) and gives $8\pi$ as its result.