# Calculus II Practice (2)

2017-07-25 15:42:18 +0000 -

Evaluate the integral.

Q1) $\int_{0}^{4\pi} \int_{0}^{5\pi} (sin \ x + cos \ y) dx dy$

A1)

Q2) $\int_{1}^{3} \int_{0}^{y} x^{2}y^{2} dx dy$

A2)

Q3) $\int_{0}^{1} \int_{0}^{y} e^{x+y} dx dy$

A3)

 \therefore \frac{1}{2}(e-1)^2

Change the Cartesian integral to an equivalent polar integral, and then evaluate.

Q4) $\int_{-7}^{7} \int_{0}^{ \sqrt{49-x^{2}} } dy dx$

A4)

$-7 \leq x \leq 7$, $0 \leq y \leq \sqrt{49-x^{2}}$

Q5) $\int_{-7}^{0} \int_{- \sqrt{49-x^{2}} }^{0} \frac{1}{ 1 + \sqrt{x^{2}+y^{2}} }dy dx$

A5)

$-7 \leq x \leq 0$, $-\sqrt{49-x^{2}} \leq y \leq 0$

$0 \leq r \leq 7$, $\pi \leq \theta \leq \frac{3}{2}\pi$

Q6) $\int_{-4}^{4} \int_{- \sqrt{16-y^{2}} }^{0} \frac{ \sqrt{x^{2}+y^{2}}}{ 1 + \sqrt{x^{2}+y^{2}} }dx dy$

A6)

$- \sqrt{16-y^{2}} \leq x \leq 0$, $-4 \leq y \leq 4$

Use the given transformation to evaluate the integral.

Q21) $u = x + y$, $v = -2x + y$

where $R$ is the parallelogram bounded by the lines $y = -x + 1$, $y = -x + 4$, $y = 2x + 2$, $y = 2x + 5$

A21)

$u = x + y$, $v = -2x + y$

$1 \leq u \leq 4$, $2 \leq v \leq 5$

$\frac{3}{2}y = u + \frac{1}{2}v$, $y = \frac{2}{3}u + \frac{1}{3}v$

If r(t) is the position vector of a particle in the plane at time t, find the indicated vector.

Q24) Find the velocity vector.

A24)

Velocity Vector:

$x'(t) = -6t$, $y'(t) = \frac{1}{7}t^{2}$

Q25) Find the acceleration vector.

A25)

Acceleration Vector:

The position vector of a particle is r(t). Find the requested vector.

Q26) The velocity at $t = 4$ for $r(t) = (9t^{2} + 3t + 10)i - 8t^{3}j + (5 - t^{2})k$

A26)

Q27) The acceleration at $t = 2$ for $r(t) = (3t - 3t^{4})i + (6 - t)j + (2t^{2} - 5t)k$

A27)

Find the unit tangent vector of the given curve.

Q28) $r(t) = 4t^{6}i - 3t^{6}j + 12t^{6}k$

A28)

Q29) $r(t) = (4 - 2t)i + (2t - 4)j + (9 + t)k$

A29)

Calculate the arc length of the indicated portion of the curve $r(t)$.

Q30) $r(t) = 10t^{3}i + 11t^{3}j + 2t^{3}k$, $-1 \leq t \leq 2$

A30)

Evaluate the line integral of $f(x,y)$ along the curve C.

Q35) $f(x,y)= \frac{x^{5}}{ \sqrt{1+4y} }$, $C: y = x^{2}$, $0 \leq x \leq 3$

A35)

Q36) $f(x, y) = x^{2} + y^{2}$, $C: y = -2x - 4$, $0 \leq x \leq 3$

A36)

Evaluate the line integral along the curve C.

Q37) $\int_{C} (y+z) ds$, $C$ is the straight-line segment $x = 0$, $y = 3 - t$, $z = t$ from $(0, 3, 0)$ to $(0, 0, 3)$

A37)

Apply Green’s Theorem to evaluate the integral.

Q41) $\oint_{c} (5y + x) dx + (y + 4x) dy$

$C$: The circle $(x - 2)^{2} + (y - 3)^{2} = 4$

A41)

Q42) $\oint_{c} (y^{2} + 3) dx + (x^{2} + 6) dy$

$C$: The triangle bounded by $x = 0$, $x + y = 5$, $y = 0$

A42)

$0 \leq x \leq 5$, $0 \leq y \leq -x+5$

Find the divergence of the field F.

Q43) $F = 15xz^{2}i + 4yj - 5z^{3}k$

A43)

Divergence $\overrightarrow{F} = 15z^{2} +4 -15z^{2}$

Evaluate the surface integral of G over the surface S.

Q46) $S$ is the plane $x + y + z = 2$ above the rectangle $0 \leq x \leq 3$ and $0 \leq y \leq 2$; $G(x,y,z) = 4z$

A46)