I am very new to integrals. If someone would kindly take a look at them and confirm they are set up correctly that would be great!
- D is the triangle with vertices $(0,0) (4,-2) (4,8)$ Evaluate $\int \int_D e^{x^2 +1}dA$
My attempt:
$\int_0^4 \int_0^{2x} e^{x^2 +1}dydx$ + $\int_0^4 \int_0^{\frac{1}{2}x} e^{x^2 +1}dydx$
$\approx 18116212.53$
- D is the region where $x \geq 0$ bounded by $z = 4-x^2-y^2$ and xy plane. Evaluate $\int \int \int_D z+2dA$
My attempt:
$\int_\frac{-\pi}{2}^{\frac{\pi}{2}} \int_0^2 \int_0^{4-r^2} (z+2)r dzdrd\theta$
$= \frac{40\pi}{3}$
I believe your second one is correct but your first integral should be set up like so:
$$\int_0^4 \int_{-.5x}^0 e^{x^2+1} dydx + \int_0^4 \int_{0}^{2x} e^{x^2+1} dydx$$
$$\approx 30193687.5441$$
If someone else wants to verify....