Vector Calculus

Question

How do we evaluate the volume under the surface given by, z=2xsin(y) over the region bounded below by the curve y=x^2 and above by the line y=0 for 0<=x<=1

Answer 1

To solve this, let z = f(x,y). This makes it clear that you are integrating with respect to x and y. The order of integration is determined by the relationships between x and y. In this case, we can write y = f(x), so would want to do the integration with respect to y first. If it were done in the reverse, you would have to consider the fact that y is a function of x, so would be much more difficult.

You are told that the function y = x² is the lower limit and y = 0 is the upper limit (one is below, the other above). The values of x are between 0 and 1; so these are the lower and upper limits of x, respectively.

This leads to the integral:

∫₀¹ ∫⁰_x^2 (2x siny) dy dx

since x does not depend on y, can separate it out:

∫₀¹ 2x [∫⁰_x^2 ( sin (y}) dy] dx

The integral of ∫ sin(u) du = - cos{u) + C

so ∫⁰_x^2 ( sin (y}) dy = ( - cos(y)|_x^2⁰ = - cos (0) - (- cos(x²)) = cos(x²) - 1

You now have an integral in x only: ∫₀¹ 2x [ cos(x²) - 1 ] dx, which can be separated into: ∫₀¹ cos(x²) 2x dx - ∫₀¹ 2x dx,

The second part of this is a simple integral you should be able to solve easily. For the first part, apply the identity:

∫ cos(u) du = sin{u) + C, where u = x² and du = 2x the limits are unchanged because u = 1² = 1 when x = 1 and u = 0² = 0 when x = 0

Once you have solved both parts, combine them and that should give you the answer.