Question

Evaluate x2 dv, e where e is the solid that lies within the cylinder x2 + y2 = 4, above the plane z = 0, and below the cone z2 = 49x2 + 49y2.

Answer 1

Answer:

$\int_{e}x^2dv = \frac{49\cdot 2^6\cdot \pi }{6}=\frac{1568\pi}{3}$

Step-by-step explanation:

We want to evaluate the $\int_{e}x^2dv$, where e the solid that lies between the cylinder $x^2+y^2=4$ and $0\leq z \leq 49(x^2+y^2)$.

To calculate this integral, we will use cylindric coordinates. They are given by the formulas $x=r\cos(\theta), y = r\sin(\theta), z=z$, where r is the distance from the point (x,y,0) to the origin and theta is the angle that the point (x,y,0) forms with the x-axis.

We will describe the solid e in this coordinates. Recall that in this coordinates $r^2 = x^2+y^2$. Then, the boundary $x^2+y^2=4$ is $r^2=4$. Also, the boundary $z=49(x^2+y^2)$ becomes $z=49r^2$. Then, we know that $0\leq z\leq r^2$. We need to define the ranges of r and theta to fully describe the solid e. Recall that since e is bounded by the cylinder, which is circular-alike, we need to have $0\leq \theta \leq 2\pi$ to describe the whole cylinder. Also, notice that the distance of any point of the form (x,y,0), where (x,y,z) is part of the solid e, lays inside the circle of equation

Now we need to express the original integral in terms of this new coordinates. To guarantee that it keeps the same result, we need the multiply the function by the jacobian of the change of coordinates. For the cylindric coordinates, the jacobian is r. Then

$\int_{e}x^2dv =\int_{0}^{2\pi}\int_{0}^{2}\int_{0}^{49r^2}r(r^2\cos^2(\theta))dzdrd\theta$. By solving the integral with respect to z, we have that

$\int_{e}x^2dv = \int_{0}^{2\pi}\int_{0}^{2} 49 r^5 \cos^2(\theta) dr d\theta$. By solving the integral with respect to r, we get

$\int_{e}x^2dv = \frac{49\cdot 2^6}{6}\int_{0}^{2\pi}\cos^2(\theta)$

By solving the final integral, we get

$\int_{e}x^2dv = \frac{49\cdot 2^6\cdot \pi }{6}=\frac{1568\pi}{3}$